Question 12 Marks
Simplify the following: $\frac{(4\times10^7)(6\times10^5)}{8\times10^4}$
Answer$\frac{(4\times10^7)(6\times10^{-4})}{8\times10^4}$ $\frac{(4\times10^7)(6\times10^5)}{8\times10^4}$ $\frac{(24\times10^7\times10^{-5})}{8\times10^4}$ $=\frac{24\times10^{7-5}}{8\times10^4}$ $=\frac{24\times10^{2}}{8\times10^4}$ $=\frac{(3\times10^2)}{10^4}$$=\frac{3}{100}$
View full question & answer→Question 22 Marks
Solve the following equations: $3^{\text{x}-1}\times5^{2\text{y}-3}=225$
Answer$3^{\text{x}-1}\times5^{2\text{y}-3}=225$ $\Rightarrow3^{\text{x}-1}\times5^{2\text{y}-3}=9\times25$ $\Rightarrow3^{\text{x}-1}\times^{2\text{y}-3}=3^2\times5^2$ $\Rightarrow\text{x}-1=2$ and $2\text{y}=3=2$ $\Rightarrow\text{x}=3$ and $\text{y}=\frac{5}{2}$
View full question & answer→Question 32 Marks
If $\text{x}=2^{\frac{1}{3}}+2^{\frac{2}{3}},$ show that $\text{x}^3-6\text{x}=6$
Answer$\text{x}=2^{\frac{1}{3}}+2^{\frac{2}{3}},$ $\Rightarrow\text{x}^3=\big(2^{\frac{1}{3}}+2^{\frac{2}{3}}\big)^3$ $\Rightarrow\text{x}^3=\Big(2^{\frac{1}{3}}\Big)^3+\Big(2^{\frac{2}{3}}\Big)^3+3\times2^{\frac{1}{3}}\times2^\frac{2}{3}\Big(2^\frac{1}{3}+2^\frac{2}{3}\Big)$ $\Rightarrow\text{x}^3=2+2^2+3\times2\Big(2^\frac{1}{3}+2^\frac{2}{3}\Big)$ $\Rightarrow\text{x}^3=6+6\Big(2^{\frac{1}{3}}+2^\frac{2}{3}\Big)$ $\Rightarrow\text{x}^3=6+6\text{x}$ $\text{x}^3-6\text{x}=6$
View full question & answer→Question 42 Marks
Show that:$\frac{1}{1+\text{x}^{\text{a}+\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}}=1$
Answer$\text{LHS}=-\frac{1}{1+\text{x}^{\text{a}+\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}}=1$ Multiplying the numerators and denominators of two terms on $L.H.S.$
by $x^b$ and $x^a$ respectively, we obtain
$\text{LHS}=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^{\text{a}-\text{b}+\text{b}}}+\frac{\text{x}^\text{a}}{\text{x}^{\text{a}}+\text{x}^{\text{b}-\text{a}+\text{a}}}$
$=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^\text{a}}+\frac{\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}^\text{b}}$
$=\frac{\text{x}^\text{b}+\text{x}^\text{a}}{\text{x}^\text{b}+\text{x}^\text{a}}$
$=1$
$=\text{RHS}$
View full question & answer→Question 52 Marks
Solve the following equations for $x:$
$2^{\text{x}+1}=4^{\text{x}-3}$
AnswerWe have,$=2^{\text{x}+1}=4^{\text{x}-3}$
$=2^{\text{x}+1}=2^{\text{2x}-6}$
$=\text{x}+1=2\text{x}-6$
$=\text{x}=7$
View full question & answer→Question 62 Marks
Solve the following equations for $x:$
$2^{3\text{x}-7}=256$
AnswerWe have,$2^{3\text{x}-7}=256$
$2^{3\text{x}-7}=2^8$
$3\text{x}-7=8$
$3\text{x}=15$
$\text{x}=5$
View full question & answer→Question 72 Marks
Simplify the following: $\frac{5\times25^{\text{n}+1}-25\times5^{2\text{n}}}{5\times5^{2\text{n}+3}-(25)^{\text{n}+1}}$
Answer$\frac{5\times25^{\text{n}+1}-25\times5^{2\text{n}}}{5\times5^{2\text{n}+3}-(25)^{\text{n}+1}}$ $=\frac{(5\times25^\text{n}\times25)-(25\times25)^\text{n}}{(5\times25^\text{n}\times125)(25)^\text{n}\times25}$ $=\frac{25^\text{n}\times25(5-1)}{25^\text{n}\times25(25-1)}$ $=\frac{4}{24}=\frac{1}{6}$
View full question & answer→Question 82 Marks
Assuming that $x, y, z$ are positive real numbers, simplify the following: $\sqrt{\text{x}^{3}\text{y}^{-2}}$
AnswerWe have, $\sqrt{\text{x}^{3}\text{y}^{-2}}=\sqrt{\frac{\text{x}^3}{\text{y}^2}}$
$=\Big(\frac{\text{x}^2}{\text{y}^2}\Big)^{\frac{1}{2}}$
$=\frac{\text{x}^{3\times\frac{1}{2}}}{\text{y}^{2\times\frac{1}{2}}}$
$=\frac{\text{x}^{\frac{3}{2}}}{\text{y}}$
$\Rightarrow\sqrt{\text{x}^3\text{y}^{-2}}=\frac{\text{x}^{\frac{3}{2}}}{\text{y}}$
View full question & answer→Question 92 Marks
Simplify the following: $\frac{6(8)^{\text{n}+1}+16(2)^{3\text{n}-2}}{10(2)^{3\text{n}+1}-7(8)^\text{n}}$
Answer$\frac{(6\times8^{\text{n}+1})+(16\times2^{3\text{n}-2})^{}}{(10\times2^{3\text{n}+1})-7\times(8)^{\text{n}}}$ $=\frac{(6\times8^\text{n}\times8)+(16\times8^\text{n}\times\frac{1}{4})}{(10\times8^\text{n}\times2)-7\times(8)^\text{n}}$ $=\frac{8^\text{n}(48+4)}{8^\text{n}(20-7)}$ $=\frac{52}{13}=4$
View full question & answer→Question 102 Marks
Simplify the following: $\frac{(\text{a}^{3\text{n-9}})^6}{\text{a}^{2\text{n}-4}}$
Answer$\frac{(\text{a}^{3\text{n-9}})^6}{\text{a}^{2\text{n}-4}}$ $=\frac{\text{a}^{18\text{n}-54}}{\text{a}^{2\text{n-4}}}=\text{a}^{18\text{n}-2\text{n}-54 +4}$ $=\text{a}^{18\text{n}-2\text{n}-54+4}=\text{a}^{16\text{n}-50}$
View full question & answer→Question 112 Marks
Simplify the following: $\frac{3\text{n}\times9^{\text{n}+1}}{3^{\text{n}-1}\times9^{\text{n}-1}}$
Answer$\frac{3\text{n}\times9^{\text{n}+1}}{3^{\text{n}-1}\times9^{\text{n}-1}}$ $=\frac{3^\text{n}\times9^\text{n}\times9}{\frac{3\text{n}}{3}\times\frac{9\text{n}}{9}}=9\times3\times9=243$
View full question & answer→Question 122 Marks
Solve the following equations for x: $2^{5\text{x}+3}=8^{\text{x}+3}$
AnswerWe have,$=2^{5\text{x}+3}=8^{\text{x}+3}$
$=2^{5\text{x}+3}=2^{\text{3x}+9}$
$=5\text{x}+3=3\text{x}+9$
$=2\text{x}=6$
$=\text{x}=3$
View full question & answer→Question 132 Marks
Write the value of $\sqrt[3]{7}\times\sqrt[3]{49}.$
AnswerWe have to find the value of $\sqrt[3]{7}\times\sqrt[3]{49}.$ So,
$\sqrt[3]{7}\times\sqrt[3]{49}=\sqrt[3]{7}\times\sqrt[3]{49}$
$=7^{\frac{1}{3}}\times7^{2\times\frac{1}{3}}$
$=7^{\frac{1}{3}}\times7^{\frac{2}{3}}$
By using law rational exponents $\text{a}^{\text{m}}\times\text{a}^\text{n}=\text{a}^{\text{m}+\text{n}}$ we get,
$\sqrt[3]{7}\times\sqrt[3]{49}=7^{\frac{1}{3}}\times7^{\frac{2}{3}}$
$=7^{\frac{1}{3}+\frac{2}{3}}$
$=7^\frac{{1+2}}{3}$
$=7^{\frac{3}{3}}=7$
Hence the value of $\sqrt[3]{7}\times\sqrt[3]{49}$ is $7.$
View full question & answer→Question 142 Marks
Show that: $\Big(\frac{\text{x}^{\text{a}^2+\text{b}^2}}{\text{x}^{\text{ab}}}\Big)^{\text{a}+\text{b}}\Big(\frac{\text{x}^{\text{b}^2+\text{c}^2}}{\text{x}^\text{bc}}\Big)^{\text{b}+\text{c}}\Big(\frac{\text{x}^{\text{c}^2+\text{a}^2}}{\text{x}^{\text{ac}}}\Big)^{\text{a}+\text{c}}=\text{x}^{2(\text{a}^2+\text{b}^2+\text{c}^2)}$
Answer$\text{LHS}=\Big(\frac{\text{x}^{\text{a}^2+\text{b}^2}}{\text{x}^{\text{ab}}}\Big)^{\text{a}+\text{b}}\Big(\frac{\text{x}^{\text{b}^2+\text{c}^2}}{\text{x}^\text{bc}}\Big)^{\text{b}+\text{c}}\Big(\frac{\text{x}^{\text{c}^2+\text{a}^2}}{\text{x}^{\text{ac}}}\Big)^{\text{a}+\text{c}}$
$=\big(\text{x}^{\text{a}^2+\text{b}^2-\text{ab}}\big)^{\text{a}+\text{b}}\big(\text{x}^{\text{b}^2+\text{c}^2-\text{bc}}\big)^{\text{b}+\text{c}}\big(\text{x}^{\text{c}^2+\text{a}^\text{2}-\text{ac}}\big)^{\text{a}+\text{c}}$
$=\big(\text{x}^{(\text{a}+\text{b})(\text{a}^2+\text{b}^2-\text{ab})}\big)\big(\text{x}^{(\text{b}+\text{c})(\text{b}^2+\text{c}^2-\text{bc})}\big)\big(\text{x}^{(\text{a}+\text{c})(\text{a}^2+\text{c}^2-\text{ac})}\big)$
$=\big(\text{x}^{\text{a}^3-\text{b}^1}\big)\big(\text{x}^{\text{b}^1-\text{c}^1}\big)\big(\text{x}^{\text{c}^1-\text{a}^1}\big)$
$=\text{x}^{\text{a}^1-\text{b}^1+\text{b}^1-\text{c}^1+\text{c}^1-\text{a}^1}$
$=\text{x}^0$
$=1$
$=\text{RHS}$
View full question & answer→Question 152 Marks
Write $\Big(\frac{1}{9}\Big)^{-\frac{1}{2}}\times(64)^{-\frac{1}{3}}$ as a rational number.
AnswerWe have to find the value of $\Big(\frac{1}{9}\Big)^{-\frac{1}{2}}\times(64)^{-\frac{1}{3}}$
So, $\Big(\frac{1}{9}\Big)^{-\frac{1}{2}}\times(64)^{-\frac{1}{3}}=\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}$
$=\Big(\frac{1}{3^2}\Big)^{\frac{-1}{2}}\times(4^3)^{\frac{-1}{3}}$
$=\Bigg(\frac{1}{3^{2\text{x}\frac{-1}{2}}}\Bigg)\times\Big(4^{3\times\frac{-1}{3}}\Big)$
$\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}=\frac{1}{3^{-1}}\times4^{-1}$
$=\frac{1}{\frac{1}{3}}\times\frac{1}{4}$
$=1\times\frac{3}{1}\times\frac{1}{4}$
$=\frac{3}{4}$ Hence the value of the value of $\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}$ is $\frac{3}{4}.$
View full question & answer→Question 162 Marks
Solve the following equations for x: $4^{2\text{x}}=\frac{1}{32}$
AnswerWe have,$=4^{2\text{x}}=\frac{1}{32}$
$=2^{4\text{x}}=\frac{1}{2}^5$
$=2^{4\text{x}}=2^{-5}$
$=4\text{x}=-5$
$=\text{x}=-\frac{5}{4}$
View full question & answer→Question 172 Marks
Simplify the following: $\frac{5^{\text{n}+3}-6\times5^{\text{n+1}}}{9\times5^{\text{x}}-2^2\times5^{\text{n}}}$
Answer$\frac{(5^{\text{n}+3})-(6\times5^{\text{n}+1})}{(9\times5^{\text{n}})-(2^2\times5^\text{n})}$
$=\frac{(5^{\text{n}+3})-(6\times5^{\text{n}+1})}{(9\times5^{\text{n}})-(2^2\times5^{\text{n}})}$
$=\frac{(5^\text{n}\times5^3)-(6\times5^\text{n}\times5)}{(9\times5^\text{n})-(2^2\times\text{5}^\text{n})}$
$=\frac{5^\text{n}(125-30)}{5^{\text{n}(9-4)}}$
$=\frac{95}{5}=19$
View full question & answer→Question 182 Marks
Write $(625)^{-\frac{1}{4}}$ in decimal form.
AnswerWe have to write $(625)^{\frac{-1}{4}}$ in decimal form.
So, $(625)^{\frac{-1}{4}}=\frac{1}{625^{\frac{1}{4}}}$
$=\Big(\frac{1}{(5^4)}\Big)^{\frac{1}{4}}$
$(625^{\frac{-1}{4}})=\Big(\frac{1}{5}\Big)^{4\times\frac{1}{4}}$
$=\frac{1}{5}$
$=0.2$ Hence the decimal form of $(625)^{-\frac{1}{4}}$ is $0.2$
View full question & answer→Question 192 Marks
State the power law of exponents.
AnswerThe "power rule" tell us that to raise a power to a power, just multiply the exponents.
If $a$ is any real number and $m, n$ are positive integers, then $\left(a^m\right)^n=a^{m n}$
We have,
$\left(a^m\right)^n=a^m \times a^m \times a^m \times \ldots n$ factors
$\left(a^m\right)^n=(a \times a \times a \times \ldots m) \times(a \times a \times a \times \ldots m) \ldots n$ factors
$\left(a^m\right)^n=(a \times a \times a \times \ldots m n)$
Hence, $\left(a^m\right)^n=a^{m n}$
View full question & answer→Question 202 Marks
If $5^{3\text{x}}=125$ and $10^{\text{y}}=0.001$ find $x$ and $y.$
Answer$5^{3\text{x}}=125$
$\Rightarrow5^{3\text{x}}=5^3$
$\Rightarrow3\text{x}=3$
$\Rightarrow\text{x}=1$ And, $10^\text{y}=0.001$
$\Rightarrow10^\text{y}=\frac{1}{1000}$
$\Rightarrow10^\text{y}=\frac{1}{10^3}$
$\Rightarrow10^\text{y}=10^{-3}$
$\Rightarrow\text{y}=-3$ Hence, $\text{x}-1$ and $\text{y}=-3$
View full question & answer→Question 212 Marks
Solve the following equations for x: $7^{2\text{x}+3}=1$
AnswerWe have,$\Rightarrow7^{2\text{x}+3}=1$
$\Rightarrow7^{2\text{x}+3}=7^0$
$\Rightarrow2\text{x}+3=0$
$\Rightarrow2\text{x}=-3$
$\Rightarrow\text{x}=-\frac{3}{2}$
View full question & answer→Question 222 Marks
Assuming that $x, y, z$ are positive real numbers, simplify the following:
$\sqrt[5]{243\text{x}^{10}\text{y}^5\text{z}^{10}}$
AnswerWe have,
$\sqrt[5]{243\text{x}^{10}\text{y}^5\text{z}^{10}}=(243\text{x}^{10}\text{y}^5\text{z}^{10})^{\frac{1}{5}}$
$=(243)^{\frac{1}{5}}\times\text{x}^{\frac{10}{5}}\text{y}^{\frac{5}{5}}\text{z}^{\frac{10}{5}}$
$=(3^5)^{\frac{1}{5}}\text{x}^2\text{y}^1\text{z}^2$
$=3^{5\times\frac{1}{5}}\text{x}^2\text{y}^2$
$=3\text{x}^2\text{yz}^2$
$\Rightarrow\sqrt[5]{243\text{x}^{10}\text{y}^5\text{z}^{10}}=3\text{x}^2\text{yz}^2$
View full question & answer→Question 232 Marks
Simplify:$\sqrt[5]{(32)^{-3}}$
AnswerWe have,$\sqrt[5]{(32)^{-3}}=[(32)^{-1}]^{\frac{1}{5}}$
$=[(2^5)^{-3}]^\frac{1}{5}$
$=2^{5\times(-3)\times\frac{1}{5}}$
$=2^{-3}$
$=\frac{1}{2^3}$
$=\frac{1}{8}$
View full question & answer→Question 242 Marks
Solve the following equations: $3^{\text{x}+1}=27\times3^4$
Answer$3^{\text{x}+1}=27\times3^4$
$\Rightarrow3^{\text{x}+1}=3^3\times3^4$
$\Rightarrow3^{\text{x}+1}+3^7$
$\Rightarrow\text{x}+1=7$
$\Rightarrow\text{x}=6$
View full question & answer→Question 252 Marks
Simplify the following: $\Big(\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}\Big)^\text{n}$
Answer$\Big(\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}\Big)^\text{n}$ $=\frac{\text{x}^{2\text{n}}\text{y}^{2\text{n}}}{\text{a}^{2\text{n}}\text{b}^{3\text{n}}}$
View full question & answer→Question 262 Marks
Find the values of x in each of the following:$\big(13\big)^{\sqrt{\text{x}}}=4^4-3^4-6$
Answer$\big(13\big)^{\sqrt{\text{x}}}=4^4-3^4-6$$\Rightarrow13^\sqrt{\text{x}}=265-81-6$
$\Rightarrow13^{\sqrt{3}}=169$
$\Rightarrow13^{\sqrt{\text{x}}}=13^2$
$\Rightarrow\sqrt{\text{x}}=2$
$\Rightarrow\text{x}=4$
View full question & answer→Question 272 Marks
Assuming that $x, y, z$ are positive real numbers, simplify the following: $\Big(\text{x}^{-\frac{2}{3}}\text{y}^{-\frac{1}{2}}\Big)^2$
AnswerWe have, $\Big(\text{x}^{-\frac{2}{3}}\text{y}^{-\frac{1}{2}}\Big)^2=\bigg(\frac{1}{\text{x}^{\frac{2}{3}\times2}\text{y}^{2\times\frac{1}{2}}}\bigg)$
$\bigg(\frac{1}{\text{x}^{\frac{2}{3}\times2}\text{y}^{2\times\frac{1}{2}}}\bigg)$
$=\frac{1}{\text{x}^{\frac{4}{3}}\text{y}^{1}}$
$=\frac{1}{\text{x}^{\frac{4}{3}}\text{y}}$
$\Rightarrow\Big(\text{x}^{-\frac{2}{3}}\text{y}^{-\frac{1}{2}}\Big)^2=\frac{1}{\text{x}^{\frac{4}{3}}\text{y}}$
View full question & answer→Question 282 Marks
If $\text{a}=\text{x}^{\text{m}+\text{n}}\text{y}^\text{l},\ \text{b}=\text{x}^{\text{n}+\text{l}}\text{y}^\text{m}$ and $\text{c}=\text{x}^{\text{l}+\text{m}\text{y}^\text{n}},$ prove that $\text{a}^{\text{m}-\text{n}}\text{b}^{\text{n}-\text{l}}\text{c}^{\text{l}-\text{m}}=1.$
Answer$\text{a}=\text{x}^{\text{m}+\text{n}}\text{y}^\text{l},\ \text{b}=\text{x}^{\text{n}+\text{l}}\text{y}^\text{m}$ and $\text{c}=\text{x}^{\text{l}+\text{m}\text{y}^\text{n}},$
Now, $\text{LHS}=\text{a}^{\text{m}-\text{n}}\text{b}^{\text{n}-\text{l}}\text{c}^{\text{l}-\text{m}}$
$=\big(\text{x}^{\text{m}+\text{n}}\text{y}^\text{l}\big)^{\text{m}-\text{n}}\times\big(\text{x}^{\text{n}+\text{l}}\big)^{\text{n}-\text{l}}\times\big(\text{x}^{\text{l}+\text{m}}\text{y}^\text{n}\big)^{\text{l}-\text{m}}$
$=\text{x}^{(\text{m}+\text{n})(\text{m}-\text{n})}\times\text{y}^{\text{l}(\text{m}-\text{n})}\times\text{x}^{(\text{n}+\text{l})(\text{n}-\text{l})}\times\text{y}^{\text{m}(\text{n}-\text{l})}\times\text{x}^{(\text{l}+\text{m})(\text{l}-\text{m})}\times\text{y}^{\text{n}(\text{l}-\text{m})}$
$=\text{x}^{\text{m}^2-\text{n}^2}\times\text{y}^{\text{lm}-\text{nl}}\times\text{x}^{\text{n}^2-\text{l}^2}\times\text{y}^{\text{mn}-\text{lm}}\times\text{x}^{\text{l}^2-\text{m}^2}\times\text{y}^{\text{nl}-\text{mn}}$
$=\text{x}^{\text{m}^2-\text{n}^2+\text{n}^2-\text{l}^2+\text{l}^2-\text{m}^2}\times\text{y}^{\text{lm}-\text{nl}+\text{mn}-\text{lm}+\text{nl}-\text{mn}}$
$=\text{x}^0\times\text{y}^0$
$=1\times1$
$=1$
$=\text{RHS}$
View full question & answer→Question 292 Marks
Assuming that $x, y, z $are positive real numbers, simplify the following: $\Big(\frac{\text{x}^{-4}}{\text{y}^{-10}}\Big)^{\frac{5}{4}}$
AnswerWe have, $\Big(\frac{\text{x}^{-4}}{\text{y}^{-10}}\Big)^{\frac{5}{4}}=\Big(\frac{\text{y}^{10}}{\text{x}^4}\Big)^\frac{5}{4}$
$=\frac{\text{y}^{10\times\frac{5}{4}}}{\text{x}^{4\times\frac{5}{4}}}$
$=\frac{\text{y}^{\frac{25}{2}}}{\text{x}^5}$
$\Rightarrow\Big(\frac{\text{x}^{-4}}{\text{y}^{-10}}\Big)^{\frac{5}{4}}=\frac{\text{y}^{\frac{25}{2}}}{\text{x}^5}$
View full question & answer→Question 302 Marks
Simplify:
$(0.001)^{\frac{1}{3}}$
Answer We have, $(0.001)^{\frac{1}{3}}=\Big(\frac{1}{1000}\Big)^{\frac{1}{3}}$
$=\Big(\frac{1}{10^3}\Big)^{\frac{1}{3}}$
$=\frac{1}{10^{3\times\frac{1}{3}}}$
$=\frac{1}{10}$
$=0.1$
$\Rightarrow0.001^{\frac{1}{3}}=0.1$
View full question & answer→Question 312 Marks
Assuming that $x, y, z$ are positive real numbers, simplify the following: $(\sqrt{\text{x}^{-3}})^5$
AnswerWe have, $(\sqrt{\text{x}^{-3}})^5=\Big(\sqrt{\frac{1}{\text{x}^3}}\Big)^5$
$=\bigg(\frac{1}{\text{x}^{\frac{3}{2}}}\bigg)^5$
$=\frac{1}{\text{x}^{\frac{3}{5}\times5}}$
$=\frac{1}{\text{x}^{\frac{15}{2}}}$
$\Rightarrow(\sqrt{\text{x}^{-3}})^{5}=\frac{1}{\text{x}^{\frac{15}{2}}}$
View full question & answer→Question 322 Marks
If $\text{x}=\text{a}^{\text{m}+\text{n}},\ \text{y}=\text{a}^{\text{n}+\text{l}}$ and $\text{z}=\text{a}^{\text{l}+\text{m}},$ prove that $\text{x}^\text{m}\text{y}^\text{n}\text{z}^\text{l}=\text{x}^\text{n}\text{y}^\text{l}\text{z}^\text{m}.$
Answer$\text{x}=\text{a}^{\text{m}+\text{n}},\ \text{y}=\text{a}^{\text{n}+\text{l}}$ and $\text{z}=\text{a}^{\text{l}+\text{m}},$
$\text{LHS}=\text{x}^\text{m}\text{y}^\text{n}\text{z}^\text{l}$
$=\big(\text{a}^{\text{m}+\text{n}}\big)^\text{m}\times\big(\text{a}^{\text{n}+\text{l}}\big)^{\text{n}}\times\big(\text{a}^{\text{l}+\text{m}}\big)^\text{l}$
$=\text{a}^{\text{m}^{2}+\text{mn}}\times\text{a}^{\text{n}^2+\text{nl}}\times\text{a}^{\text{l}^2+\text{lm}}$
$=\text{a}^{\text{m}^2+\text{mn}+\text{n}^2+\text{nl}+\text{l}^2+\text{lm}}$
$=\text{a}^{\text{mn}+\text{n}^2}\times\text{a}^{\text{nl}+\text{l}^2}\times\text{a}^{\text{lm}+\text{m}^2}$
$=\text{a}^{\text{n}(\text{m}+\text{n})}\times\text{a}^{\text{l}(\text{n}+\text{l})}\times\text{a}^{\text{m}(\text{l}+\text{m})}$
$=\big(\text{a}^{(\text{m}+\text{n})}\big)^\text{n}\times\big(\text{a}^{(\text{n}+\text{l})}\big)^\text{l}\times\big(\text{a}^{(\text{l}+\text{m})}\big)^\text{m}$
$=\text{x}^\text{n}\text{y}^\text{l}\text{z}^\text{m}$
View full question & answer→Question 332 Marks
Solve the following equations: $\sqrt{\frac{\text{a}}{\text{b}}}=\Big(\frac{\text{b}}{\text{a}}\Big)^{1-2\text{x}},$ where a, b are distinct positive primes.
Answer$\sqrt{\frac{\text{a}}{\text{b}}}=\Big(\frac{\text{b}}{\text{a}}\Big)^{1-2\text{x}}$
$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)^{\frac{1}{2}}=\Big(\frac{\text{b}}{\text{a}}\Big)^{1-2\text{x}}$
$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)^{\frac{1}{2}}=\Big(\frac{\text{a}}{\text{b}}\Big)^{-1+2\text{x}}$
$\Rightarrow\frac{1}{2}=-1+2\text{x}$
$\Rightarrow2\text{x}=\frac{1}{2}+1$
$\Rightarrow2\text{x}=\frac{3}{2}$
$\Rightarrow2\text{x}=\frac{3}{4}$
View full question & answer→Question 342 Marks
Solve the following equations:
$4^{\text{x}-1}\times(0.5)^{3-2\text{x}}=\Big(\frac{1}{8}\Big)^{\text{x}}$
Answer $4^{\text{x}-1}\times(0.5)^{3-2\text{x}}=\Big(\frac{1}{8}\Big)^{\text{x}}$
$\Rightarrow(2^2)^{\text{x}-1}\times\Big(\frac{1}{2}\Big)^{3-2\text{x}}=\Big(\frac{1}{2^3}\Big)^\text{x}$
$\Rightarrow2^{2\text{x}-2}\times2^{-3+2\text{x}}=2^{-3\text{x}}$
$\Rightarrow2^{2\text{x}-2-3+2\text{x}}=2^{-3\text{x}}$
$\Rightarrow2^{4\text{x}-5}=2^{-3\text{x}}$
$\Rightarrow4\text{x}-5=-3\text{x}$
$\Rightarrow7\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{7}$
View full question & answer→Question 352 Marks
Write the value of $\sqrt[3]{125\times27}.$
AnswerWe have to find the value of $\sqrt[3]{125\times27}.$
So, $\sqrt[3]{125\times27}=\sqrt[3]{5^3\times3^3}=5\times3=15$
Hence the value of the value of $\sqrt[3]{125\times27}$ is $15.$
View full question & answer→Question 362 Marks
If $2^4 \times 4^2=16^x$, then find the value of $x$.
AnswerWe have to find the value of $x$ provided $2^4 \times 4^2=16^x$
So, $2^4 \times\left(2^2\right)^2=16^x 2^4 \times 2^4=2^{4 x} 2^{4+4}=2^{4 x}$
By equating the exponents we get $4+4=4 \times 8=4 \times \frac{8}{4}=x 2=x$
Hence the value of $x$ is $2 .$
View full question & answer→Question 372 Marks
Show that: $\frac{\Big(\text{a}+\frac{1}{\text{b}}\Big)^\text{m}\times\Big(\text{a}-\frac{1}{\text{b}}\Big)^\text{n}}{\Big(\text{b}+\frac{1}{\text{a}}\Big)^\text{m}\times\Big(\text{b}-\frac{1}{\text{a}}\Big)^\text{n}}=\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}+\text{n}}$
Answer$\text{LHS}=\frac{\Big(\text{a}+\frac{1}{\text{b}}\Big)^\text{m}\times\Big(\text{a}-\frac{1}{\text{b}}\Big)^\text{n}}{\Big(\text{b}+\frac{1}{\text{a}}\Big)^\text{m}\times\Big(\text{b}-\frac{1}{\text{a}}\Big)^\text{n}}$
$=\frac{\Big(\frac{\text{ab}+1}{\text{b}}\Big)^\text{m}\Big(\frac{\text{ab}-1}{\text{b}}\Big)^\text{n}}{\Big(\frac{\text{ab}+1}{\text{a}}\Big)^\text{m}\Big(\frac{\text{ab}-1}{\text{a}}\Big)^{\text{n}}}$
$=\frac{\frac{\text{(ab}+1)^\text{m}}{\text{b}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{b}^\text{n}}}{\frac{(\text{ab}+1)^\text{m}}{\text{a}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{a}^\text{n}}}$
$=\frac{(\text{ab}+1)^\text{m}}{\text{b}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{b}^\text{n}}\times\frac{\text{a}^\text{m}}{(\text{ab}+1)^\text{m}}\times\frac{\text{a}^\text{n}}{(\text{ab}-1)^\text{n}}$
$=\frac{\text{a}^{\text{m}+\text{n}}}{\text{b}^{\text{m}+\text{n}}}$
$=\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}+\text{n}}$
$=\text{RHS}$
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Find the values of $x$ in each of the following:$\Big(\sqrt{\frac{3}{5}}\Big)^{\text{x}+1}=\frac{125}{27}$
Answer$\Big(\sqrt{\frac{3}{5}}\Big)^{\text{x}+1}=\frac{125}{27}$$\Rightarrow\Big(\frac{3}{5}\Big)^{\frac{1}{2}\times(\text{x}+1)}=\frac{5^3}{3^3}$
$\Rightarrow\Big(\frac{3}{5}\Big)^{\frac{\text{x}+1}{2}}=\Big(\frac{5}{3}\Big)^3$
$\Rightarrow\Big(\frac{3}{5}\Big)^{\frac{\text{x}+1}{2}}=\Big(\frac{3}{5}\Big)^{-3}$
$\Rightarrow\frac{\text{x}+1}{2}=-3$
$\Rightarrow\text{x}+1=-6$
$\Rightarrow\text{x}=-7$
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If $1176=2^\text{a}\times3^\text{b}\times7^\text{c},$ find $a, b$ and $c.$
AnswerGiven that $2, 3$ and $7$ are factors of $1176.$
Taking out the $LCM$ of $1176$,
we get $2^3\times3^1\times7^2=2^\text{a}\times3^\text{b}\times7^\text{c}$
By commparing, we get
$\text{a}=3,\ \text{b}=1$ and $\text{c}=2.$
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Simplify:$\Big(16^{-\frac{1}{5}}\Big)^{\frac{5}{2}}$
AnswerWe have,$\Big(16^{-\frac{1}{5}}\Big)^{\frac{5}{2}}=16^{-\frac{1}{5}\times\frac{5}{2}}$
$=16^{-\frac{1}{2}}$
$=\frac{1}{16^{\frac{1}{2}}}$
$=\frac{1}{(4^{2})^{\frac{1}{2}}}$
$=\frac{1}{(4^2)^{\frac{1}{2}}}$
$=\frac{1}{4}$
$\Rightarrow\Big(16^{-\frac{1}{5}}\Big)^{\frac{5}{2}}=\frac{1}{4}$
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Simplify the following: $\frac{4\text{ab}^2(-5\text{ab}^3)}{10\text{a}^2\text{b}^2}$
Answer$\frac{4\text{ab}^2(-5\text{ab}^3)}{10^2\text{b}^2}$ $=-\frac{20\text{a}^2\text{b}^5}{10\text{a}^2\text{b}^2}=-2\text{b}^3$
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Simplify:$\sqrt[3]{(343)^{-2}}$
AnswerWe have,$\sqrt[3]{(343)^{-2}}=\sqrt[3]{\frac{1}{(343)^2}}$
$=\frac{1}{(343)^{\frac{2}{3}}}$
$=\frac{1}{(7^3)^{\frac{2}{3}}}$
$=\frac{1}{7^{3\times\frac{2}{3}}}$
$=\frac{1}{7^2}$
$=\frac{1}{49}$
$\Rightarrow\sqrt[3]{(343)^{-2}}=\frac{1}{49}$
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If a and b are distinct primes such that $\sqrt[3]{\text{a}^6\text{b}^{-4}}=\text{a}^\text{x}\text{b}^{2\text{y}},$find x and y.
Answer$\sqrt[3]{\text{a}^6\text{b}^{-4}}=\text{a}^\text{x}\text{b}^{2\text{y}}$
$\Rightarrow(\text{a}^6\text{b}^{-4})^\frac{1}{3}=\text{a}^\text{x}\times\text{b}^{2\text{y}}$
$\Rightarrow\text{a}^{6\times\frac{1}{3}}\times\text{b}^{-\frac{4}{3}}=\text{a}^\text{x}\times\text{b}^{2\text{y}}$
$\Rightarrow\text{a}^2\times\text{b}^{-\frac{4}{3}}=\text{a}^\text{x}\times\text{b}^{2\text{y}}$
$\Rightarrow2=\text{x}$ and $-\frac{4}{3}=2\text{y}$
$\Rightarrow\text{x}=2$ and $\text{y}=-\frac{2}{3}$
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Show that: $\big(\text{x}^{\text{a}-\text{b}}\big)^{\text{a}+\text{b}}\big(\text{x}^{\text{b}-\text{c}}\big)^{\text{b}+\text{c}}\big(\text{x}^{\text{c}-\text{a}}\big)^{\text{c}+\text{a}}=1$
Answer$\text{LHS}=\big(\text{x}^{\text{a}-\text{b}}\big)^{\text{a}+\text{b}}\big(\text{x}^{\text{b}-\text{c}}\big)^{\text{b}+\text{c}}\big(\text{x}^{\text{c}-\text{a}}\big)^{\text{c}+\text{a}}$
$=\big(\text{x}^{(\text{a}+\text{b})(\text{a}+\text{b})}\big)\big(\text{x}^{(\text{b}-\text{c})(\text{b}+\text{c})}\big)\big(\text{x}^{(\text{c}-\text{a})(\text{c}+\text{a})}\big)$
$=\big(\text{x}^{\text{a}^2-\text{b}^2}\big)\big(\text{x}^{\text{b}^2-\text{c}^2}\big)\big(\text{x}^{\text{c}^2-\text{a}^2}\big)$
$=\text{x}^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$=\text{x}^0$
$=1$
$=\text{RHS}$
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Solve the following equations for x: $4^{\text{x}-1}\times(0.5)^{3-2\text{x}}=\Big(\frac{1}{8}\Big)^\text{x}$
AnswerWe have,$4^{\text{x}-1}\times(0.5)^{3-2\text{x}}=\Big(\frac{1}{8}\Big)^\text{x}$
$2^{2\text{x}-2}\times\Big(\frac{1}{2}\Big)^{3-2\text{x}}=\Big(\frac{1}{2}\Big)^{3\text{x}}$
$2^{2\text{x}-2}\times2^{2\text{x}-3}=\Big(\frac{1}{2}\Big)^{3\text{x}}$
$2^{2\text{x}-2+2\text{x}-3}=\Big(\frac{1}{2}\Big)^{3\text{x}}$
$2^{4\text{x}-5}=2^{-3\text{x}}$
$4\text{x}-5=-3\text{x}$
$7\text{x}=5$
$\text{x}=\frac{5}{7}$
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Show that:$\Big(\frac{3^\text{a}}{2^{\text{b}}}\Big)^{\text{a}+\text{b}}\Big(\frac{3^\text{b}}{3^\text{c}}\Big)^{\text{b}+\text{c}}\Big(\frac{3^\text{c}}{3^\text{a}}\Big)^{\text{c}+\text{a}}=1$
Answer$\text{LHS}=\Big(\frac{3^\text{a}}{2^{\text{b}}}\Big)^{\text{a}+\text{b}}\Big(\frac{3^\text{b}}{3^\text{c}}\Big)^{\text{b}+\text{c}}\Big(\frac{3^\text{c}}{3^\text{a}}\Big)^{\text{c}+\text{a}}$$=\big(3^{\text{a}-\text{b}}\big)^{\text{a}+\text{b}}\big(3^{\text{b}-\text{c}}\big)^{\text{b}+\text{c}}\big(3^{\text{c}-\text{a}}\big)^{\text{c}+\text{a}}$
$=\big(3^{(\text{a}-\text{b})(\text{a}+\text{b})}\big)\big(3^{(\text{b}-\text{c})(\text{b}+\text{c})}\big)\big(3^{(\text{c}-\text{a})(\text{c}+\text{a})}\big)$
$=\big(3^{\text{a}^2-\text{b}^2}\big)\big(3^{\text{b}^2-\text{c}^2}\big)\big(3^{\text{c}^2-\text{a}^2}\big)$
$=3^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$=3^0$
$=1$
$=\text{RHS}$
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