Question 13 Marks
Simplify: $\Big(\frac{\sqrt{2}}{5}\Big)^8\div\Big(\frac{\sqrt{2}}{5}\Big)^{13}$
AnswerWe have, $\Big(\frac{\sqrt{2}}{5}\Big)^8\div\Big(\frac{\sqrt{2}}{5}\Big)^{13}$
$=\Big(\frac{\sqrt{2}}{5}\Big)^8\times\Big(\frac{5}{\sqrt{2}}\Big)^{13}$
$=\frac{(\sqrt{2})^8}{5^8}\times\frac{5^{13}}{(\sqrt{2})^{13}}$
$=\frac{5^{13}\times5^{-8}}{(\sqrt{2})^{13}\times(\sqrt{2})^{-8}}$
$=\frac{5^{13-8}}{(\sqrt{2})^{13-8}}$
$=\frac{5^5}{(\sqrt{2})^5}=\frac{3125}{4\sqrt{2}}$
$\Rightarrow\Big(\frac{\sqrt{2}}{5}\Big)\div\Big(\frac{\sqrt{2}}{5}\Big)^{13}=\frac{3125}{4\sqrt{2}}$
View full question & answer→Question 23 Marks
Prove that: $\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\frac{1}{\Big(\frac{256}{625}\Big)^\frac{1}{4}}+\Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0=\frac{61}{16}$
Answer$\text{LHS}=\Big(\frac{64}{125}\Big)^{-\frac{2}{3}}+\frac{1}{\Big(\frac{256}{625}\Big)^\frac{1}{4}}+\Big(\frac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0=\frac{61}{16}$
$=\Big(\frac{4^3}{5^3}\Big)^{\frac{-2}{3}}+\frac{1}{\Big(\frac{4^4}{5^4}\Big)^{\frac{1}{4}}}+1$
$=\frac{4^{3\times\Big(\frac{-2}{3}\Big)}}{5^{3\times\Big(\frac{-2}{3}\Big)}}+\frac{1}{\frac{4^{4\times\frac{1}{4}}}{5^{4\times\frac{1}{4}}}}+1$
$=\frac{4^{-2}}{5^{-2}}+\frac{1}{\frac{4}{5}}+1$
$=\frac{5^2}{4^2}+\frac{5}{4}+1$
$=\frac{25+20+16}{16}$
$=\frac{61}{16}$
$=\text{RHS}$
View full question & answer→Question 33 Marks
Show that:$\Big(\frac{\text{a}^{\text{x}+1}}{\text{a}^{\text{y}+1}}\Big)^{\text{x}+\text{y}}\Big(\frac{\text{a}^{\text{y}+2}}{\text{a}^{\text{z}+2}}\Big)^{\text{y}+\text{z}}\Big(\frac{\text{a}^{\text{z}+3}}{\text{a}^{\text{x}+3}}\Big)^{\text{z}+\text{x}}=1$
Answer $\text{LHS}=\Big(\frac{\text{a}^{\text{x}+1}}{\text{a}^{\text{y}+1}}\Big)^{\text{x}+\text{y}}\Big(\frac{\text{a}^{\text{y}+2}}{\text{a}^{\text{z}+2}}\Big)^{\text{y}+\text{z}}\Big(\frac{\text{a}^{\text{z}+3}}{\text{a}^{\text{x}+3}}\Big)^{\text{z}+\text{x}}$$=\big(\text{a}^{\text{x}+1-(\text{y}+1)}\big)^{\text{x}+\text{y}}\big(\text{a}^{\text{y}+2-(\text{x}+2)}\big)^{\text{y}+\text{z}}\big(\text{a}^{\text{z}+3-(\text{x}+3)}\big)^{\text{z}+\text{x}}$
$=\big(\text{a}^{\text{x}+1-\text{y}-1}\big)^{\text{x}+\text{y}}\big(\text{a}^{\text{y}+2-\text{z}-2}\big)^{\text{y}+\text{z}}\big(\text{a}^{\text{z}+3-\text{x}-3}\big)^{\text{z}+\text{x}}$
$=\big(\text{a}^{\text{x}-\text{y}}\big)^{\text{x}+\text{y}}\big(\text{a}^{\text{y}-\text{z}}\big)^{\text{y}+\text{z}}\big(\text{a}^{\text{z}-\text{x}}\big)^{\text{z}+\text{x}}$
$=\big(\text{a}^{(\text{x}-\text{y})(\text{x}+\text{y})}\big)\big(\text{a}^{(\text{y}-\text{z})(\text{y}+\text{z})}\big)\big(\text{a}^{(\text{z}-\text{x})(\text{z}+\text{x})}\big)$
$=\big(\text{a}^{\text{x}^2-\text{y}^2}\big)\big(\text{a}^{\text{y}^2-\text{z}^2}\big)\big(\text{a}^{(\text{z}^2-\text{x}^2)}\big)$
$=\text{a}^{\text{x}^2-\text{y}^2+\text{y}^2-\text{z}^2+\text{z}^2-\text{x}^2}$
$=\text{a}^0$
$=1$
$=\text{RHS}$
View full question & answer→Question 43 Marks
If $3^{4\text{x}}=(81)^{-1}$ and $10^{\frac{1}{\text{y}}}0.0001,$ find the value of $2^{-\text{x}+4\text{y}}.$
Answer$3^{4\text{x}}=(81)^{-1}$
$\Rightarrow3^{4\text{x}}=\frac{1}{81}$
$\Rightarrow3^{4\text{x}}=\frac{1}{3^4}$
$\Rightarrow3^{4\text{x}}=3^{-4}$
$\Rightarrow4\text{x}=-4$
$\Rightarrow\text{x}=-1$ And, $10^\frac{1}{\text{y}}=0.0001$
$\Rightarrow10^\frac{1}{\text{y}}=\frac{1}{10000}$
$\Rightarrow10^\frac{1}{\text{y}}=\frac{1}{10^4}$
$\Rightarrow10^\frac{1}{\text{y}}=10^{-4}$
$\Rightarrow\frac{1}{\text{y}}=-4$
$\Rightarrow\text{y}=-\frac{1}{4}$
$\therefore2^{-\text{x}+4\text{y}}=2^{-(-1)+4\text{x}\Big(-\frac{1}{4}\Big)}=2^{1-1}=2^0=1$
View full question & answer→Question 53 Marks
Prove that: $\frac{2^\text{n}+2^{\text{n-1}}}{2^{\text{n}+1}-2^\text{n}}=\frac{3}{2}$
AnswerWe have, $\frac{2^\text{n}+2^{\text{n-1}}}{2^{\text{n}+1}-2^\text{n}}=\frac{2^\text{n}+2^\text{n}\times2^{-1}}{2^\text{n}\times\text{2}^1-2^\text{n}}$
$=\frac{2^\text{n}[1+2^{-1}]}{2^\text{n}[2-1]}$
$=\frac{1+\frac{1}{2}}{1}$
$=1+\frac{1}{2}$
$=\frac{3}{2}$
$\Rightarrow\frac{2^\text{n}+2^{\text{n}-1}}{2^{\text{n}+1}-2^\text{n}}=\frac{3}{2}$
View full question & answer→Question 63 Marks
State the product law of exponents.
AnswerState the product law of exponents. If $a$ is any real number and $m, n$ are positive integers, then $a^m \times a^n=a^{m+n}$ By definition, we have $a^m \times a^n=(a \times a \times \ldots m$ factor $) \times(a \times a \times \ldots n$ factor $) a^m \times a^n=a \times a \ldots$ to $(m+n)$ factors $a^m \times a^n$
$=a^{m+n}$ Thus the exponent "product rule" tells us that, when multiplying two powers that have the same base, we can add the exponents.
View full question & answer→Question 73 Marks
Find the values of x in each of the following:$5^{\text{x}-2}\times3^{2\text{x}-3}=135$
AnswerWe have,$5^{\text{x}-2}\times3^{2\text{x}-3}=135$
$\Rightarrow5^{\text{x}-2}\times3^{2\text{x}-3}=5\times3^3$
On equating the exponents, we get
We get,
$\text{x}-2=1$ and $2\text{x}-3=3$
$\Rightarrow\text{x}=3$
Hence, $\text{x}=3$
View full question & answer→Question 83 Marks
If $2^\text{x}\times3^\text{y}\times5^\text{z}=2160,$ find $x , y $and $z$. Hence, compute the value of $3^\text{x}\times2^{-\text{y}}\times5^{-\text{z}}.$
Answer$2^\text{x}\times3^\text{y}\times5^\text{z}=2160,$ By prime factorisation, we have $2160=2\times2\times2\times2\times3\times3\times3\times5=2^4\times3^3\times5$
$\Rightarrow2^\text{x}\times3^\text{y}\times5^\text{z}=2^4\times3^3\times5$
$\Rightarrow\text{x}=4,\ \text{y}=3$ and $\text{z}=1$
$\therefore\text{3}^\text{x}\times2^{-\text{y}}\times\text{5}^{-\text{z}}=3^\text{x}\times\frac{1}{2\text{y}}\times\frac{1}{5^\text{z}}$
$=3^4\times\frac{1}{2^3}\times\frac{1}{5}$
$=81\times\frac{1}{8}\times\frac{1}{5}$
$=\frac{81}{40}$
View full question & answer→Question 93 Marks
Prove that: $\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
AnswerWe have, $\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
$=\frac{1}{2}+\frac{1}{(0.001)^{\frac{1}{2}}}-(3^3)^{\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{(0.1)^{2\times\frac{1}{2}}}-3^{3\times\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{0.1}-3^2$
$=\frac{1}{2}+10-9$
$=\frac{1}{2}+1=\frac{3}{2}$
$\Rightarrow\sqrt{\frac{1}{4}}(0.001)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
View full question & answer→Question 103 Marks
Prove that: $\Big(\frac{\text{x}^\text{a}}{\text{x}^\text{b}}\Big)^{\text{a}^2+\text{ab}+\text{b}^2}\times\Big(\frac{\text{x}^2}{\text{x}^\text{c}}\Big)^{\text{b}^2+\text{bc}+\text{c}^2}\times\Big(\frac{\text{x}^\text{c}}{\text{x}^\text{a}}\Big)^{\text{c}^2+\text{ca}+\text{a}^2}=1$
AnswerTo prove$\Big(\frac{\text{x}^\text{a}}{\text{x}^\text{b}}\Big)^{\text{a}^2+\text{ab}+\text{b}^2}\times\Big(\frac{\text{x}^2}{\text{x}^\text{c}}\Big)^{\text{b}^2+\text{bc}+\text{c}^2}\times\Big(\frac{\text{x}^\text{c}}{\text{x}^\text{a}}\Big)^{\text{c}^2+\text{ca}+\text{a}^2}=1$
Left hand side (LHS) = Right hand side (RHS) Considering LHS,$=\frac{\text{x}^{\text{a}^3}+\text{a}^2\text{b}+\text{ab}^2}{\text{x}^{\text{a}^2}{\text{b}}+\text{ab}^2+\text{b}^3}\times\frac{\text{x}^{\text{b}^3}+\text{b}^2\text{c}+\text{bc}^2}{\text{x}^{\text{b}^2}\text{c}+\text{bc}^2+\text{c}^3}\times\frac{\text{x}^{\text{c}^3}+\text{c}^2\text{a}+\text{ca}^2}{\text{x}^{\text{c}^2}\text{a}+\text{ca}^2+\text{a}^3}$
$=\text{x}^{\text{a}^3}+\text{a}^2\text{b}+\text{ab}^2-(\text{b}^3+\text{a}^2\text{b}\text{ab}^2)\\\times\text{x}^{\text{b}^3}+\text{b}^2\text{c}+\text{b}\text{c}^2-(\text{c}^3+\text{b}^2\text{c}+\text{bc}^2)\times\text{x}^{\text{c}^3}$
$+\text{c}^2\text{a}+\text{ca}^2-(\text{a}^3+\text{c}^2\text{a}+\text{ca}^2)$
$=\text{x}^{\text{a}^3}-\text{b}^3\times\text{x}^{\text{b}^3-\text{c}^3}\times\text{x}^{}\text{c}^3-\text{a}^3$
$=\text{x}^{\text{a}^3}-\text{b}^3+\text{b}^3-\text{c}^3-\text{a}^3$
$=\text{x}^0$
$=1$
Or, Therefore, $LHS = RHS$ Hence proved
View full question & answer→Question 113 Marks
Prove that: $9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=15$
AnswerWe have to prove that $9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=15$
So, $9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=3^{2\times\frac{3}{2}}-3\times5^0-\frac{1}{81^{-\frac{1}{2}}}$
$=3^{2\times\frac{3}{2}}-3\times1-\frac{1}{\frac{1}{\sqrt{81}}}$
$9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=3^3-3-\frac{1}{\frac{1}{\sqrt[2]{9\times9}}}$
$=27-3-\frac{1}{\frac{1}{9}}$
$=27-3-1\times\frac{9}{1}$
$=27-12=15$ Hence, $9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=15$
View full question & answer→Question 123 Marks
If $3^{\text{x}+1}=9^{\text{x}-2},$ find the value of $2^{1+\text{x}}.$
Answer$3^{\text{x}+1}=9^{\text{x}-2}$
$\Rightarrow3^\text{x}\times3=9^\text{x}\times9^{-2}$
$\Rightarrow3^\text{x}\times3=(3^2)^\text{x}\times\frac{1}{9^2}$
$\Rightarrow3^\text{x}\times3=3^{2\text{x}}\times\frac{1}{(3^2)^2}$
$\Rightarrow3^\text{x}\times3\times3^{2\text{x}}\times\frac{1}{3^4}$
$\Rightarrow\frac{3^{2\text{x}}}{3^\text{x}}=3\times3^4$
$\Rightarrow3^{2\text{x}-\text{x}}=3^5$
$\Rightarrow3^\text{x}=3^5$
$\Rightarrow\text{x}=5$
$\Rightarrow2^{1+\text{x}}=2^{1+5}=2^6=64$
View full question & answer→Question 133 Marks
Prove that: $\frac{\text{abc}}{\text{a}^{-1}\text{b}^{-1}+\text{b}^{-1}\text{c}^{-1}+\text{c}^{-1}\text{a}^{-1}}=\text{abc}$
Answer$\frac{\text{abc}}{\text{a}^{-1}\text{b}^{-1}+\text{b}^{-1}\text{c}^{-1}+\text{c}^{-1}\text{a}^{-1}}=\text{abc}$
Left hand side $(LHS)$ = Right hand side $(RHS)$
Considering $LHS$, $=\frac{\text{a}+\text{b}+\text{c}}{\frac{1}{\text{ab}}+\frac{1}{\text{bc}}+\frac{1}{\text{ca}}}$
$=\frac{\frac{\text{a}+\text{b}+\text{c}}{\text{a}+\text{b}+\text{c}}}{\text{abc}}$
$=\text{abc}$ Therefore, $LHS = RHS$ Hence proved
View full question & answer→Question 143 Marks
Show that:$\Big[\Big\{\frac{\text{x}^{\text{a}(\text{a}-\text{b})}}{\text{x}^{\text{a}(\text{a}+\text{b})}}\Big\}\div\Big\{\frac{\text{x}^{\text{b}(\text{b}-\text{a})}}{\text{x}^{\text{b}(\text{b}+\text{a})}}\Big\}\Big]^{\text{a}+\text{b}}=-\frac{3}{2}$
Answer$\Big[\Big\{\frac{\text{x}^{\text{a}(\text{a}-\text{b})}}{\text{x}^{\text{a}(\text{a}+\text{b})}}\Big\}\div\Big\{\frac{\text{x}^{\text{b}(\text{b}-\text{a})}}{\text{x}^{\text{b}(\text{b}+\text{a})}}\Big\}\Big]^{\text{a}+\text{b}}=1$$\text{LHS}=\Big[\Big\{\frac{\text{x}^{\text{a}(\text{a}-\text{b})}}{\text{x}^{\text{a}(\text{a}+\text{b})}}\Big\}\div\Big\{\frac{\text{x}^{\text{b}(\text{b}-\text{a})}}{\text{x}^{\text{b}(\text{b}+\text{a})}}\Big\}\Big]^{\text{a}+\text{b}}$
$\Big[\frac{\text{x}^{\text{a}(\text{a}-\text{b})}}{\text{x}^{\text{a}(\text{a}+\text{b})}}\times\frac{\text{x}^{\text{b}(\text{b}-\text{a})}}{\text{x}^{\text{b}(\text{b}+\text{a})}}\Big]^{\text{a}+\text{b}}$
$=\Big[\frac{\text{x}^{\text{a}^2-\text{ab}}}{\text{x}^{\text{a}^2-\text{ab}}}\times\frac{\text{x}^{\text{b}^2+\text{ab}}}{\text{x}^{\text{b}^2-\text{ab}}}\Big]^{\text{a}+\text{b}}$
$=\Big[\frac{\text{x}^{\text{a}^2-\text{ab}+\text{b}^2+\text{ab}}}{\text{x}^{\text{a}^2+\text{ab}+\text{b}^2-\text{ab}}}\Big]^{\text{a}+\text{b}}$
$=\Big[\frac{\text{x}^{\text{a}^2+\text{b}^2}}{\text{x}^{\text{a}^2+\text{b}^2}}\Big]^{\text{a}+\text{b}}$
$=[1]^{\text{a}+\text{b}}$
$=1$
$=\text{RHS}.$
View full question & answer→Question 153 Marks
Prove that:
$\frac{(0.6^0)-(0.1)^{-1}}{\Big(\frac{3}{8}\Big)^{-1}\Big(\frac{3}{2}\Big)^{^3}+\Big(\frac{1}{3}\Big)^{-1}}=-\frac{3}{2}$
Answer We have,
$\frac{(0.6^0)-(0.1)^{-1}}{\Big(\frac{3}{8}\Big)^{-1}\Big(\frac{3}{2}\Big)^{^3}+\Big(\frac{1}{3}\Big)^{-1}}=-\frac{3}{2}$
$=\frac{1-\frac{1}{0.1}}{\Big(\frac{8}{3}\Big)\Big(\frac{3}{2}\Big)^3+(-3)^1}$
$=\frac{1-10}{\frac{8}{3}\times\frac{3^3}{2^3}-3}$
$=\frac{-9}{3^2-3}$
$=\frac{-9}{0-3}=-\frac{9}{6}=-\frac{3}{2}$
View full question & answer→Question 163 Marks
Solve the following equations:
$4^{2\text{x}}=(\sqrt[3]{16})^{\frac{-6}{\text{y}}}=(\sqrt{8})^2$
Answer $4^{2\text{x}}=\big(\sqrt[3]{16}\big)^{\frac{-6}{\text{y}}}=\big(\sqrt{8}\big)^2$ | Consider, $4^2=\big(\sqrt{8}\big)^2$ | Now, consider, $\big(\sqrt[3]{16}\big)^{\frac{-6}{?\text{y}?}}=\big(\sqrt{8}\big)^2$ |
| $\Rightarrow\big(2^2\big)^{2\text{x}}=\Big(\sqrt{2^3}\Big)^2$ | $\Rightarrow\Big(\sqrt[3]{2^4}\Big)^{\frac{-6}{?\text{y}?}}=\Big(\sqrt{2^3}\Big)^2$ |
| $\Rightarrow2^{4\text{x}}=\Big(2^{3\times\frac{1}{2}}\Big)$ | $\Rightarrow\Big(2^{4\times\frac{1}{3}}\Big)^{\frac{-6}{\text{y}}}=\Big(2^{3\times\frac{1}{2}}\Big)^2$ |
| $\Rightarrow2^{4\text{x}}=2^{3\times\frac{1}{2}\times2}$ | $\Rightarrow2^{4\times\frac{1}{3}\times\Big(-\frac{6}{\text{y}}\Big)}=2^{3\times\frac{1}{2}\times2}$ |
| $\Rightarrow2^{4\text{x}}=2^3$ | $\Rightarrow2^{-\frac{8}{\text{y}}}=2^3$ |
| $\Rightarrow4\text{x}=3$ | $\Rightarrow-\frac{8}{\text{y}}=3$ |
| $\Rightarrow\text{x}=\frac{3}{4}$ | $\Rightarrow\text{y}=-\frac{8}{3}$ |
Hence, $\text{x}=\frac{3}{4}$and $\text{y}=-\frac{8}{3}$
View full question & answer→Question 173 Marks
If $a$ and $b$ are different positive primes such that $\Big(\frac{\text{a}^{-1}\text{b}^2}{\text{a}^2\text{b}^2}\Big)^7\div\Big(\frac{\text{a}^3\text{b}^{-5}}{\text{a}^{-1}\text{b}^3}\Big)=\text{a}^\text{x}\text{b}^\text{y},$ find $x$ and $y$.
Answer$\Big(\frac{\text{a}^{-1}\text{b}^2}{\text{a}^2\text{b}^2}\Big)^7\div\Big(\frac{\text{a}^3\text{b}^{-5}}{\text{a}^{-1}\text{b}^3}\Big)=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(\text{a}^{-1-2}\times\text{b}^{2+4})^7\div(\text{a}^{3+2}\times\text{b}^{-5-3})=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(\text{a}^{-3}\times\text{b}^{6})\div(\text{a}^5\times\text{b}^{-\text{8}})=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(\text{a}^{-21}\times\text{b}^{42})\div(\text{a}^5\times\text{b}^{-8})=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\frac{\text{a}^{21}\times\text{b}^{42}}{\text{a}^5\times\text{b}^{-8}}=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\text{a}^{-21-5}\times\text{b}^{42+8}=\text{a}^\text{x}\times\text{b}^\text{y}$
$\Rightarrow\text{a}^{-26}\times\text{b}^{50}=\text{a}^\text{x}\times\text{b}^\text{y}$
$\Rightarrow\text{x}=-26$ and $\text{y}=50$
View full question & answer→Question 183 Marks
Show that:$\Bigg\{\Big(\text{x}^{\text{a}-\text{a}^{-1}}\Big)^{\frac{1}{\text{a}+\text{1}}}\Bigg\}^{\frac{\text{a}}{\text{a}+\text{1}}}=\text{x}$
Answer$\text{LHS}=\Bigg\{\Big(\text{x}^{\text{a}-\text{a}^{-1}}\Big)^{\frac{1}{\text{a}+\text{1}}}\Bigg\}^{\frac{\text{a}}{\text{a}+\text{1}}}$$=\bigg(\text{x}^{\text{a}-\frac{1}{\text{a}}}\bigg)^{\frac{1}{\text{a}-1}\times\frac{\text{a}}{\text{a}+1}}$
$\bigg(\text{x}^{\frac{\text{a}^2-1}{\text{a}}}\bigg)^{\frac{\text{a}}{(\text{a}-1)(\text{a}+1)}}$
$=\bigg(\text{x}^{\frac{\text{a}^2-1}{\text{a}}}\bigg)^{\frac{\text{a}}{\text{a}^2-1}}$
$=\text{x}^{\frac{\text{a}^2-1}{\text{a}}\times\frac{\text{a}}{\text{a}^2-1}}$
$=\text{x}$
$=\text{RHS}$
View full question & answer→Question 193 Marks
Prove that: $\frac{1}{1+\text{x}^{\text{a}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}}=1$
Answer$\frac{1}{1+\text{x}^{\text{a}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}}=1$
Left hand side $(LHS)$ = Right hand side $(RHS)$ Considering $LHS$,
$=\frac{1}{1+\frac{\text{x}^\text{a}}{\text{x}^\text{b}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{a}}}$
$=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^\text{a}}+\frac{\text{x}^\text{a}}{\text{x}^2+\text{x}^\text{b}}$
$=\frac{\text{x}^\text{b}+\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}\text{b}}$
$=1$ Therefore, $LHS = RHS$ Hence proved
View full question & answer→Question 203 Marks
Solve the following equations for $x$: $3^{2\text{x}+4}+1=2\times3^{\text{x}+2}$
AnswerWe have,$3^{2\text{x}+4}+1=2\times3^{\text{x}+2}$
$(3^{\text{x}+2})^2+1=2\times3^{\text{x}+2}$
$\text{Let}\ 3^{\text{x}+2}=\text{y}$
$\text{y}^2+1=2\text{y}$
$\text{y}^2=2\text{y}+1=0$
$\text{y}^2-\text{y}-\text{y}+1=0$
$\text{y}(\text{y}-1)-1(\text{y}-1)=0$
$(\text{y}-1)(\text{y}-1)=0$
$\text{y}=1$
View full question & answer→Question 213 Marks
if $\text{a}=\text{xy}^{\text{p}-1},\ \text{b}=\text{xy}^{\text{q}-1}$ and $\text{c}=\text{xy}^{\text{r}-1},$ prove that $\text{a}^{\text{q}-\text{r}}\ \text{b}^{\text{r}-\text{p}}\ \text{c}^{\text{p}-\text{q}}=1.$
AnswerGiven, $\text{a}=\text{xy}^{\text{p}-1},\ \text{b}=\text{xy}^{\text{q}-1}$ and $\text{c}=\text{xy}^{\text{r}-1}$
To prove, $\text{a}^{\text{q}-\text{r}}\ \text{b}^{\text{r}-\text{q}}\ \text{c}^{\text{p}-\text{q}}=1$
Left hand side $(LHS)$ = Right hand side $(RHS)$
Considering $LHS$, $=\text{a}^{\text{q}-\text{r}}\ \text{b}^{\text{r}-\text{p}}\text{c}^{\text{p}-\text{q}}\ ...(1)$ By substituting the value of $a, b$ and $c$ in equation $(i)$, we get $= (\text{xy}^{\text{p}−1})^{\text{q}−\text{r}}(\text{xy}^{\text{q}−1})^{\text{r}−\text{p}}(\text{xy}^{\text{r}-\text{1}}) ^{\text{p}-\text{q}}$
$=\text{x}^{\text{q}-\text{r}}.\text{x}^{\text{r}-\text{p}}.\text{x}^{\text{p}-\text{q}}.\text{y}^{(\text{p}-1)\text{q}-\text{r}}.\text{y}^{(\text{q}-1)\text{r}-\text{p}}.\text{y}^{(\text{r}-1)\text{p}-\text{q}}$
$=\text{x}^{\text{q}-\text{r}+\text{r}-\text{p}+\text{p}-\text{q}}.\text{y}^{\text{pq}-\text{pr}-\text{q}+\text{r}+\text{qr}-\text{pq}-\text{r}+\text{p}+\text{rp}-\text{rq}-\text{p}+\text{q}}$
$=\text{x}^\circ.\text{y}^\circ=1$
View full question & answer→Question 223 Marks
Find the values of $x$ in each of the following:$2^{5\text{x}}\div2^\text{x}\sqrt[5]{2^{20}}$
AnswerWe have,
$2^{5\text{x}}\div2^\text{x}\sqrt[5]{2^{20}}$
$\Rightarrow\frac{2^{5\text{x}}}{2^{\text{x}}}=\big(2^{20}\big)^\frac{1}{5}$
$\Rightarrow2^{5\text{x}}\times2^{-\text{x}}=2^{20\times\frac{1}{5}}$
$\Rightarrow2^{4\text{x}}=2^4$
On equating the exponents, we get
$4\text{x}=4$
$\Rightarrow\text{x}=\frac{4}{4}=1$
Hence, $\text{x}=1$
View full question & answer→Question 233 Marks
Find the values of $x$ in each of the following:$\big(\sqrt[3]{4}\big)^{2\text{x}+\frac{1}{2}}=\frac{1}{32}$
Answer$\big(\sqrt[3]{4}\big)^{2\text{x}+\frac{1}{2}}=\frac{1}{32}$
$\Rightarrow4^{\frac{1}{3}\times\Big(2\text{x}+\frac{1}{2}\Big)}=\frac{1}{2^5}$
$\Rightarrow\big(2^2\big)^{\frac{1}{3}\times\frac{4\text{x}+1}{2}}=\frac{1}{2^5}$
$\Rightarrow2^{2\times\frac{1}{3}\times\frac{4\times+1}{2}}=\frac{1}{2^5}$
$\Rightarrow2^{\frac{4x+1}{3}}=2^{-5}$
$\Rightarrow\frac{4x+1}{3}=-5$
$\Rightarrow4\text{x}+1=-15$
$\Rightarrow4\text{x}=-16$
$\Rightarrow\text{x}=-4$
View full question & answer→Question 243 Marks
If $27^\text{x}=\frac{9}{3^\text{x}},$ find $x$.
AnswerWe have, $27^\text{x}=\frac{9}{3^\text{x}}$
$\Rightarrow\big(3^3\big)^\text{x}=\frac{3^2}{3^\text{x}}$
$\Rightarrow3^{3\text{x}}=\frac{3^2}{3^\text{x}}$
$\Rightarrow3^{3\text{x}}\times3^\text{x}=\text{3}^2$
$\Rightarrow3^{3\text{x}}+\text{x}=3^2$
$\Rightarrow3^{4\text{x}}=3^2$
On equating the exponents, we get $4\text{x}=2$
$\Rightarrow\text{x}=\frac{2}{4}=\frac{1}{2}$
Hence, $\text{x}=\frac{1}{2}$
View full question & answer→Question 253 Marks
Find the values of x in each of the following: $5^{2\text{x}+3}=1$
Answer $5^{2\text{x}+3}=1$ $\Rightarrow5^{2\text{x}}\times5^3=1$
$\Rightarrow5^{2\text{x}}\times125=1$
$\Rightarrow5^{2\text{x}}=\frac{1}{125}$
$\Rightarrow5^{2\text{x}}=\frac{1}{5^3}$
$\Rightarrow5^{2\text{x}}=5^{-3}$
$\Rightarrow2\text{x}=-3$
$\Rightarrow\text{x}=-\frac{3}{2}$
View full question & answer→Question 263 Marks
If $2^\text{x}=3^\text{y}=12^\text{z},$ show that $\frac{1}{\text{z}}=\frac{1}{\text{y}}+\frac{2}{\text{x}}.$
AnswerLet $2^\text{x}=3^\text{y}=12^\text{z}=\text{k}$ T
hen, $2=\text{k}^{\frac{1}{\text{x}}},\ \text{3}=\text{k}^{\frac{1}{\text{y}}}$ and
$12=\text{k}^{\frac{1}{\text{z}}}$
Now,$12=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow2^2\times3=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow2^2\times3=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\Big(\text{k}^{\frac{1}{\text{x}}}\Big)^2\times\text{k}^\frac{1}{\text{y}}=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^\frac{2}{\text{x}}\times\text{k}^\frac{1}{\text{y}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{2}{\text{x}}+\frac{1}{\text{y}}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\frac{2}{\text{x}}+\frac{1}{\text{y}}=\frac{1}{\text{z}}$
View full question & answer→Question 273 Marks
If $3^\text{x}=5^\text{y}=(75^\text{z}),$ show that $\text{z}=\frac{\text{xy}}{2\text{x}+\text{y}}.$
Answer Let $\text{3}^\text{x}=\text{5}^\text{y}=\text{(75)}^{\text{z}}=\text{k}$
Then, $\text{3}=\text{k}^{\frac{1}{\text{x}}},\ \text{5}=\text{k}^{\frac{1}{\text{y}}}$ and $\text{75}=\text{k}^{\frac{1}{\text{z}}}$
Now,
$75=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow3\times5^2=\text{K}^{\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}}\times\Big(\text{k}^\frac{1}{\text{y}}\Big)^2=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}}\times\text{k}^{\frac{2}{\text{y}}}=\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}+\frac{2}{\text{y}}}=\text{k}^{\frac{1}{\text{z}}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{2}{\text{y}}=\frac{1}{\text{z}}$
$\Rightarrow\frac{\text{y}+2\text{x}}{\text{xy}}=\frac{1}{\text{z}}$
$\Rightarrow\text{z}=\frac{\text{xy}}{2\text{x}+\text{y}}$
View full question & answer→Question 283 Marks
Solve the following equations for $x$:
$2^{\text{2x}}-2^{\text{x}+3}+2^4=0$
AnswerWe have,$\Rightarrow2^{\text{2x}}-2^{\text{x}+3}+2^4=0$
$\Rightarrow2^2\text{x}+2^4=2^\text{x}.2^3$
$\Rightarrow\text{Let}\ 2^\text{x}=\text{y}$
$\Rightarrow\text{y}^2+2^4=\text{y}\times2^3$
$\Rightarrow\text{y}^2=8\text{y}+16=0$
$\Rightarrow\text{y}^2-4\text{y}-4\text{y}+16=0$
$\Rightarrow\text{y}(\text{y}-4)-4(\text{y}-4)=0$
$\Rightarrow\text{y}=4$
$\Rightarrow\text{x}^2=2^2$
$\Rightarrow\text{x}=2$
View full question & answer→Question 293 Marks
Find the values of x in each of the following:$\Big(\frac{3}{5}\Big)^\text{x}\Big(\frac{5}{3}\Big)^{2\text{x}}=\frac{125}{27}$
AnswerWe have, $\Big(\frac{3}{5}\Big)^\text{x}\Big(\frac{5}{3}\Big)^{2\text{x}}=\frac{125}{27}$
$\Rightarrow\Big(\frac{3}{5}\Big)^{\text{x}}\Big(\frac{5}{3}\Big)^{2\text{x}}=\frac{5^3}{3^3}$
$\Rightarrow\Big(\frac{5}{3}\Big)^{-\text{x}}\Big(\frac{5}{3}\Big)^{2\text{x}}=\Big(\frac{5}{3}\Big)^3$
$\Rightarrow\Big(\frac{5}{3}\Big)^{-\text{x}+\text{2}\text{x}}=\Big(\frac{5}{3}\Big)^3$
$\Rightarrow\Big(\frac{5}{3}\Big)^{\text{x}}=\Big(\frac{5}{3}\Big)^3$
On equating the exponents, we get Hence, $\text{x}=3$
View full question & answer→Question 303 Marks
If $2^\text{x}=3^\text{y}=6^{-\text{z}},$ Show that $\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=0.$
AnswerLet $2^\text{x}=3^\text{y}=6^{-\text{z}}=\text{k}$ Then, $2=\text{k}^{\frac{1}{\text{x}}},\ \text{3}=\text{k}^{\frac{1}{\text{y}}}$ and $6=\text{k}^{-\frac{1}{\text{z}}}$ Now,$6=\text{k}^{-\frac{1}{\text{z}}}$
$\Rightarrow2\times3=\text{k}^{-\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}}\times\text{k}^\frac{1}{\text{y}}=\text{k}^{-\frac{1}{\text{z}}}$
$\Rightarrow\text{k}^{\frac{1}{\text{x}}+\frac{1}{\text{y}}}=\text{k}^{-\frac{1}{\text{z}}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}=-\frac{1}{\text{z}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=0$
View full question & answer→Question 313 Marks
If $a$ and $b$ are different positive primes such that $(\text{a}+\text{b})^{-1}(\text{a}^{-1}+\text{b}^{-1})=\text{a}^\text{x}\text{b}^\text{y},$ find $\text{x}+\text{y}+2.$
Answer$(\text{a}+\text{b})^{-1}(\text{a}^{-1}+\text{b}^{-1})=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\frac{1}{\text{a}+\text{b}}\times\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\frac{1}{\text{a}+\text{b}}\times\frac{\text{a}+\text{b}}{\text{ab}}=\text{a}^\text{x}\text{b}^\text{x}$
$\Rightarrow\frac{1}{\text{ab}}=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(\text{ab})^{-1}=\text{a}^\text{x}\text{b}^\text{y}$ $\Rightarrow\text{a}^{-1}\times\text{b}^{-1}=\text{a}^\text{x}\times\text{b}^\text{y}$
$\Rightarrow\text{x}=-1$ and $\text{y}=-1$
$\therefore\text{x}+\text{y}+2=-1-1+2=0$
View full question & answer→Question 323 Marks
If $1176=2^\text{a}\times3^\text{b}\times7^\text{c},$ find the values of $a, b$ and $c$. Hence, compute the value of $2^\text{a}\times3^\text{b}\times7^{-\text{c}}$ as a fraction.
Answer$1176=2^\text{a}\times3^\text{b}\times7^\text{c}$ By prime factorisation,
we have $1176=2\times2\times2\times3\times7\times7=2^3\times3\times7^2$
$\Rightarrow2^3\times3\times7^2=2^\text{a}\times3^\text{b}\times7^\text{c}$
$\Rightarrow\text{a}=3,\ \text{b}=1$ and $\text{c}=2$
$\therefore2^\text{a}\times3^\text{b}\times7^{-\text{c}}=2^3\times3^1\times7^{-2}$
$=8\times3\times\frac{1}{7^2}$ $=24\times\frac{1}{49}$
$=\frac{24}{49}$
View full question & answer→Question 333 Marks
If $\text{a}^\text{x}=\text{b}^\text{y}=\text{c}^\text{z}$ and $\text{b}^2=\text{ac},$ then show that $\text{y}=\frac{2\text{zx}}{\text{z}+\text{x}}$
Answer Let $\text{a}^\text{x}=\text{b}^\text{y}=\text{c}^{\text{z}}=\text{k}$
Then, $\text{a}=\text{k}^{\frac{1}{\text{x}}},\ \text{b}=\text{k}^{\frac{1}{\text{y}}}$ and $\text{c}=\text{k}^{\frac{1}{\text{z}}}$
Now,
$\text{b}=\text{ac}$
$\Rightarrow\Big(\text{k}^\frac{1}{\text{y}}\Big)^2=\text{k}^\frac{1}{\text{x}}\times\text{k}^\frac{1}{\text{z}}$
$\Rightarrow\text{k}^\frac{2}{\text{y}}=\text{k}^{\frac{1}{\text{x}}+\frac{1}{\text{z}}}$
$\Rightarrow\frac{2}{\text{y}}=\frac{1}{\text{x}}+\frac{1}{\text{z}}$
$\Rightarrow\frac{2}{\text{y}}=\frac{\text{z}+\text{x}}{\text{zx}}$
$\Rightarrow\text{y}=\frac{2\text{zx}}{\text{z}+\text{x}}$
View full question & answer→Question 343 Marks
Prove that: $\Big(\sqrt{3\times5^{-3}}\div\sqrt[3]{3^{-1}}\sqrt{5}\times\sqrt[6]{3\times5^6}=\frac{3}{5}\Big)$
AnswerWe have to prove that $\frac{\sqrt{3\times5^{-3}}}{\sqrt[3]{3^{-1}\sqrt{5}}}\times\sqrt[6]{3\times5^6}=\frac{3}{5}$
By using rational exponent $\text{a}^{-\text{n}}=\frac{1}{\text{a}^{\text{n}}}$
we get, $\frac{\sqrt{3\times5^{-3}}}{\sqrt[3]{3^{-1}\sqrt{5}}}\times\sqrt[6]{3\times5^6}=\frac{\sqrt{3\times\frac{1}{5^3}}}{\sqrt[3]{\frac{1}{3}}\sqrt{5}}\times\sqrt[6]{3\times5^6}$
$\frac{\sqrt{3\times5^{-3}}}{\sqrt[3]{3^{-1}\sqrt{5}}}\times\sqrt[6]{3\times5^6}=\frac{3^\frac{1}{2}\times\frac{1}{5^{3\times\frac{1}{2}}}}{\frac{1}{3^{\frac{1}{3}}}\times5^\frac{1}{2}}\times3^\frac{1}{6}\times5^{6\times\frac{1}{6}}$
$=\frac{\frac{3^\frac{1}{2}}{5^{\frac{3}{2}}}}{\frac{5^{\frac{1}{5}}}{3^\frac{1}{3}}}\times3^\frac{1}{6}\times5^1$
$=\frac{3^\frac{1}{2}}{5^\frac{3}{2}}\times\frac{3^\frac{1}{3}}{5^\frac{1}{2}}\times3^\frac{1}{6}\times5^1$
$\frac{\sqrt{3\times5^{-3}}}{\sqrt[3]{3^{-1}\sqrt{5}}}\times\sqrt[6]{3\times5^6}=3^\frac{1}{3}\times3^\frac{1}{3}\times5^{-\frac{3}{2}}\times5^{-\frac{1}{2}}\times3^{\frac{1}{6}}\times5^1$
$=3^{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}\times5^{-\frac{3}{2}-\frac{1}{2}+1}$
$\frac{\sqrt{3\times5^{-3}}}{\sqrt[3]{3^{-1}\sqrt{5}}}\times\sqrt[6]{3\times5^6}=3^{\frac{1\times3}{2\times3}+\frac{1\times2}{3\times2}+\frac{1}{6}}\times5^{-\frac{3}{2}-\frac{1}{2}+\frac{1}{6}}$
$=3^{\frac{3+2+1}{6}}\times5^{\frac{-3-1+2}{2}}$
$\frac{\sqrt{3\times5^{-3}}}{\sqrt[3]{3^{-1}\sqrt{5}}}\times\sqrt[6]{3\times5^6}3^\frac{6}{6}\times5^\frac{2}{2}=3^1\times5^{-1}=\frac{3}{5}$
Hence, $\frac{\sqrt{3\times5^{-3}}}{\sqrt[3]{3^{-1}\sqrt{5}}}\times\sqrt[6]{3\times5^6}=\frac{3}{5}$
View full question & answer→Question 353 Marks
Find the values of $x$ in each of the following:$\big(2^3\big)^4=\big(2^2\big)^\text{x}$
AnswerWe have, $\big(2^3\big)^4=\big(2^2\big)^\text{x}$
$\Rightarrow2^{3\text{x}\times4}=2^{2\times\text{x}}$
$\Rightarrow2^{12}=2^2\text{x}$
On equating the exponents,we get $2\text{x}=12$
$\Rightarrow\text{x}=\frac{12}{2}=6$ Hence, $\text{x}=6$
View full question & answer→Question 363 Marks
Simplify: $\Big(\frac{5^{-1}\times7^2}{5^2\times7^{-4}}\Big)^{\frac{7}{2}}\times\Big(\frac{5^{-2}\times7^3}{5^3\times7^{-5}}\Big)^{-\frac{5}{2}}$
AnswerWe have, $\Big(\frac{5^{-1}\times7^2}{5^2\times7^{-4}}\Big)^{\frac{7}{2}}\times\Big(\frac{5^{-2}\times7^3}{5^3\times7^{-5}}\Big)^{-\frac{5}{2}}$
$=\Big(\frac{7^{2+4}}{5^{2+1}}\Big)^{\frac{7}{2}}\times\Big(\frac{7^{3+5}}{5^{3+2}}\Big)^{-\frac{5}{2}}$
$=\Big(\frac{7^6}{5^3}\Big)^{\frac{7}{2}}\times\Big(\frac{7^8}{5^5}\Big)^{-\frac{5}{2}}$
$=\frac{7^{6\times\frac{7}{2}}}{5^{3\times\frac{7}{2}}}\times\frac{7^{8\times-\frac{5}{2}}}{5^{5\times-\frac{5}{2}}}$
$=\frac{7^{21}}{5^{\frac{21}{2}}}\times\frac{7^{-20}}{5^{-\frac{25}{2}}}$
$=\frac{7^{21-20}}{5^{\frac{21}{2}-\frac{25}{2}}}=\frac{7}{5^{-\frac{4}{2}}}$
$=7\times5^{\frac{4}{2}}=7\times5^2$
$=7\times25=175$
$\Rightarrow\Big(\frac{5^{-1}\times7^2}{5^2\times7^{-4}}\Big)^{\frac{7}{2}}\times\Big(\frac{5^{-2}\times7^3}{5^3\times7^{-5}}\Big)^{-\frac{5}{2}}=175$
View full question & answer→Question 373 Marks
Write the value of $\Big\{5\Big(8^{\frac{1}{3}}+27^{\frac{1}{3}}\Big)^3\Big\}^{\frac{1}{4}}.$
AnswerWe have to find the value of $\Big\{5\Big(8^{\frac{1}{3}}+27^{\frac{1}{3}}\Big)^3\Big\}^{\frac{1}{4}}.$ So,
$\Big\{5\Big(8^{\frac{1}{3}}+27^{\frac{1}{3}}\Big)^3\Big\}^{\frac{1}{4}}=\Big\{5\Big(2^{3\times\frac{1}{3}}+3^{3\times\frac{1}{3}}\Big)\Big\}^{\frac{1}{4}}$
$=\Big\{5\Big(2^{3\times\frac{1}{3}}+3^{3\times\frac{1}{3}}\Big)\Big\}^{\frac{1}{4}}$
$=\big\{5(2+3)^3\big\}^{\frac{1}{4}}$
$=\big\{5\times5^3\big\}^{\frac{1}{4}}$
By using rational exponents $\text{a}^\text{m}\times\text{a}^\text{n}=\text{a}^{\text{m+n}}$ we get
$\Big\{5\Big(8^{\frac{1}{3}}+27^{\frac{1}{3}}\Big)^3\Big\}^{\frac{1}{4}}=\big\{5^{1+3}\big\}^{\frac{1}{4}}$
$=5^{4\times\frac{1}{4}}$
$=5^1$
$=5$
Hence the simplified value of $\Big\{5\Big(8^{\frac{1}{3}}+27^{\frac{1}{3}}\Big)^3\Big\}^{\frac{1}{4}}$ is $5$.
View full question & answer→Question 383 Marks
Prove that:
$(\text{a}^{-1}+\text{b}^{-1})=\frac{\text{ab}}{\text{a}+\text{b}}$
Answer$\text{LHS}=(\text{a}^{-1}+\text{b}^{-1})$
$=\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)^{-1}$
$=\Big(\frac{\text{b}+\text{a}}{\text{ab}}\Big)^{-1}$
$=\frac{\text{ab}}{\text{a}+\text{b}}$
$=\text{RHS}$
Therefore, $LHS = RHS$ Hence proved
View full question & answer→Question 393 Marks
Find the values of x in each of the following:$2^{\text{x}-7}\times5^{\text{x}-4}=1250$
AnswerWe have,$2^{\text{x}-7}\times5^{\text{x}-4}=1250$
$\Rightarrow2^{\text{x}-7}\times5^{\text{x}-4}=2^1\times5^4$
On equating the exponents, we get
We get,
$\text{x}-7=1$ and $\text{x}-4=4$
$\Rightarrow\text{x}=8$
Hence, $\text{x}=8$
View full question & answer→Question 403 Marks
Show that: $\Big(\frac{1}{\text{x}^{\text{a}-\text{b}}}\Big)^{\frac{1}{\text{a}-\text{c}}}\Big(\frac{1}{\text{x}^{\text{b}-\text{c}}}\Big)^{\frac{1}{\text{b}-\text{a}}}\Big(\frac{1}{\text{x}^{\text{c}-\text{a}}}\Big)^{\frac{1}{\text{c}-\text{b}}}=1$
Answer$\text{LHS}=\Big(\frac{1}{\text{x}^{\text{a}-\text{b}}}\Big)^{\frac{1}{\text{a}-\text{c}}}\Big(\frac{1}{\text{x}^{\text{b}-\text{c}}}\Big)^{\frac{1}{\text{b}-\text{a}}}\Big(\frac{1}{\text{x}^{\text{c}-\text{a}}}\Big)^{\frac{1}{\text{c}-\text{b}}}$
$=\Big(\text{x}^{\frac{1}{\text{a}-\text{b}}\times\frac{1}{\text{a}-\text{c}}}\Big)\Big(\text{x}^{\frac{1}{\text{b}-\text{c}}\times\frac{1}{\text{b}-\text{a}}}\Big)\Big(\text{x}^{\frac{1}{\text{c}-\text{a}}\times\frac{1}{\text{a}-\text{b}}}\Big)$
$=\Big(\text{x}^{\frac{1}{\text{a}-\text{b}}\times\frac{1}{\text{a}-\text{c}}+\frac{1}{\text{b}-\text{c}}\times\frac{1}{\text{b}-\text{c}}+\frac{1}{\text{c}-\text{a}}\times\frac{1}{\text{c}-\text{b}}}\Big)$
$=\bigg(\text{x}^{-\frac{1}{(\text{a}-\text{b})(\text{c}-\text{a})}-\frac{1}{(\text{b}-\text{c})(\text{a}-\text{b})}-\frac{1}{(\text{c}-\text{a})(\text{b}-\text{c})}}\bigg)$
$=\bigg(\text{x}^{\frac{-(\text{b}-\text{c})-(\text{c}-\text{a})-(\text{a}-\text{b})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}}\bigg)$
$=\bigg(\text{x}^{\frac{-\text{b}+\text{c}-\text{c}+\text{a}-\text{a}+\text{b}}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}}\bigg)$
$=\big(\text{x}^0\big)$
$=1$
$=\text{RHS}$
View full question & answer→Question 413 Marks
Assuming that $x, y, z$ are positive real numbers, simplify the following: $(\sqrt{\text{x}})^{-\frac{2}{3}}\sqrt{\text{y}^4}\div\sqrt{\text{xy}^{-\frac{1}{2}}}$
AnswerWe have, $(\sqrt{\text{x}})^{-\frac{2}{3}}\sqrt{\text{y}^4}\div\sqrt{\text{xy}^{-\frac{1}{2}}}$
$=\Big(\text{x}^{\frac{1}{2}}\Big)^{-\frac{2}{3}}(\text{y}^2)\div\sqrt{\text{xy}^{-\frac{1}{2}}}$
$=\frac{\text{x}^{\frac{1}{2}\times-\frac{2}{3}}\text{y}^2}{\Big(\text{xy}^{-\frac{1}{2}}\Big)^{\frac{1}{2}}}$
$=\frac{\text{x}^{-\frac{1}{3}}\text{y}^2}{\text{x}^{\frac{1}{2}}\text{y}^{-\frac{1}{2}\times\frac{1}{2}}}$
$=\Big(\text{x}^{-\frac{1}{3}}\times\text{x}^{-\frac{1}{2}}\Big)\times\Big(\text{y}^2\times\text{y}^{\frac{1}{4}}\Big)$
$=\Big(\text{x}^{-\frac{1}{2}-\frac{1}{2}}\Big)\Big(\text{y}^{2+\frac{1}{4}}\Big)$
$=\Big(\text{x}^{\frac{-2-3}{6}}\Big)\Big(\text{y}^{\frac{8+1}{4}}\Big)$
$=\Big(\text{x}^{-\frac{5}{6}}\Big)\Big(\text{y}^{\frac{9}{4}}\Big)$
$=\frac{\text{y}^{\frac{9}{4}}}{\text{x}^{\frac{5}{6}}}$
$\Rightarrow(\sqrt{\text{x}})^{-\frac{2}{3}}\sqrt{\text{y}^4}\div\sqrt{\text{xy}^{-\frac{1}{2}}}=\frac{\text{y}^{\frac{9}{4}}}{\text{x}^{\frac{5}{6}}}$
View full question & answer→Question 423 Marks
Assuming that $x, y, z$ are positive real numbers, simplify the following: $\Big(\frac{\sqrt{2}}{\sqrt{3}}\Big)^5\Big(\frac{6}{7}\Big)^2$
AnswerWe have, $\Big(\frac{\sqrt{2}}{\sqrt{3}}\Big)^5\Big(\frac{6}{7}\Big)^2=\frac{(\sqrt{2})^5}{(\sqrt{3})^5}\times\frac{6^2}{7^2}$
$=\frac{4\sqrt{2}}{9\sqrt{3}}\times\frac{36}{49}$
$=\frac{4\sqrt{2}\times4}{\sqrt{3}\times49}$
$=\frac{\sqrt{256}\times\sqrt{2}}{\sqrt{3}\times\sqrt{2401}}$
$=\frac{\sqrt{512}}{\sqrt{7203}}$
$=\Big(\sqrt{\frac{512}{7203}}\Big)$
$=\Big(\frac{512}{7203}\Big)^\frac{1}{2}$
View full question & answer→Question 433 Marks
If $3^{x-1}=9$ and $4^{y+2}=64$, what is the value of $\frac{x}{y}$ ?
AnswerWe have to find the value of $\frac{x}{y}$ for $3^{y+1}=9,4^{y+2}=64$
So, $3^{x-1}=3^2$ By equating the exponent
we get $x-1=2 x=2+1 x=3$
Let's take $4^{y+2}=644^{y+2}=4^3$ By equating the exponent we get .
$y+2=3 y=3-2$ By substituting $x=3, y=1$ in $\frac{x}{y}$
we get $\frac{3}{1}$ Hence the value of $\frac{x}{y}$ is $3$ .
View full question & answer→Question 443 Marks
Simplify: $\frac{(25)^{\frac{3}{2}}\times(243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}}\times(8)^{\frac{4}{3}}}$
AnswerWe have, $\frac{(25)^{\frac{3}{2}}\times(243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}}\times(8)^{\frac{4}{3}}}=\frac{(5^2)^{\frac{3}{2}}\times(3^5)^{\frac{3}{5}}}{(2^4)^{\frac{5}{4}}\times(2^3)^{\frac{4}{3}}}$
$=\frac{5^{2\times\frac{3}{2}}\times3\times^{5\times\frac{3}{2}}}{2^{4\times\frac{5}{4}}\times2^{3\times\frac{4}{3}}}$
$=\frac{5^3\times3^3}{2^5\times2^4}$
$=\frac{125\times127}{32\times16}$
$=\frac{3375}{512}$
$\Rightarrow\frac{(25)^{\frac{3}{2}}(243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}}\times(8)^{\frac{4}{3}}}=\frac{3375}{512}$
View full question & answer→Question 453 Marks
Determine $(8\text{x})^\text{x},$ if $9^{\text{x}+2}=240+9^\text{x}. $
Answer$9^{\text{x}+2}=240+9^\text{x}$
$\Rightarrow9^\text{x}\times9^2=240+9^\text{x}$
$\Rightarrow9^\text{x}\times81=240+9^\text{x}$
$\Rightarrow9^\text{x}\times81-9^\text{x}=240$
$\Rightarrow9^\text{x}(81-1)=240$
$\Rightarrow9^\text{x}\times80=240$
$\Rightarrow9^\text{x}=3$
$\Rightarrow\big(3^2\big)^\text{x}=3$
$\Rightarrow3^{2\text{x}}=3$
$\Rightarrow2\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{2}$
$\Rightarrow(8\text{x})^\text{x}=\Big(8\times\frac{1}{2}\Big)^{\frac{1}{2}}=(4)^\frac{1}{2}=2^{2\times\frac{1}{2}}=2$
View full question & answer→Question 463 Marks
Prove that: $\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}}+\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}=1$
Answer$\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}}+\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}=1$
Left hand side $(LHS)$ = Right hand side $(RHS)$ Considering $LHS$,
$=\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{a}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}}}+\frac{1}{1+\frac{\text{x}^\text{a}}{\text{x}^\text{b}}+\frac{\text{x}^\text{c}}{\text{x}^\text{b}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{c}}}$
$=\frac{\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^\text{a}+\text{x}^\text{c}}+\frac{\text{x}^\text{c}}{\text{x}^\text{c}+\text{x}^\text{b}+\text{x}^\text{a}}$
$\frac{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$
$=1$
Therefore, $LHS = RHS$ Hence proved
View full question & answer→