Question 13 Marks
If $\frac{9^n \cdot 3^2 \cdot 3^n-(27)^n}{\left(3^m \cdot 2\right)^3}=3^{-3}$Show that$: \mathrm{m}-\mathrm{n}=1$.
Answer$ \frac{9^n \cdot 3^2 \cdot 3^n-(27)^n}{\left(3^m \cdot 2\right)^3}=3^{-3}$
$ \Rightarrow \frac{3^{2 n} \cdot 3^2 \cdot 3^n-(3)^{3 n}}{3^{3 m} \cdot(2)^3}=\frac{1}{3^3}$
$ \Rightarrow \frac{3^{3 n} \cdot 3^2-3^{3 n}}{3^{3 m} \cdot 2^3}=\frac{1}{3^3}$
$ \Rightarrow \frac{3^{3 n}\left(3^2-1\right)}{3^{3 m} \times 8}=\frac{1}{3^3}$
$ \Rightarrow \frac{3^{3 n} \times 8}{3^{3 m} \times 8}=\frac{1}{3^3}$
$ \Rightarrow \frac{1}{3^{3(m-n)}}=\frac{1}{3^{3 \times 1}}$
$ \Rightarrow \mathrm{m}-\mathrm{n}=1 ($proved$) $
View full question & answer→Question 23 Marks
If $\left(\frac{a^{-1} b^2}{a^2 b^{-4}}\right)^7 \div\left(\frac{a^3 b^{-5}}{a^{-2} b^3}\right)^{-5}=a^x \cdot b^y$, find $x+y$
Answer$ \left(\frac{a^{-1} b^2}{a^2 b^{-4}}\right)^7 \div\left(\frac{a^3 b^{-5}}{a^{-2} b^3}\right)^{-5}=a^x \cdot b^y$
$ \Rightarrow\left(\frac{b^6}{a^3}\right)^7 \div\left(\frac{a^5}{b^8}\right)^{-5}=a^x \cdot b^y$
$ \Rightarrow\left(\frac{b^6}{a^3}\right)^7 \div\left(\frac{b^8}{a^5}\right)^5=a^x \cdot b^y$
$ \Rightarrow\left(\frac{b^{42}}{a^{21}}\right) \div\left(\frac{b^{40}}{a^{25}}\right)=a^x \cdot b^y$
$ \Rightarrow\left(\frac{b^{42}}{a^{21}}\right) \times\left(\frac{a^{25}}{b^{40}}\right)=a^x \cdot b^y$
$ \Rightarrow b^2 \times a^4=a^x \times b^y$
$ \Rightarrow x=4$ and $ y=2$
$ \Rightarrow x+y=4+2$
$A=6$
View full question & answer→Question 33 Marks
Simplify$ : \frac{3 \times 9^{n+1}-9 \times 3^{2 n}}{3 \times 3^{2 n+3}-9^{n+1}}$
Answer$ \frac{3 \times 9^{n+1}-9 \times 3^{2 n}}{3 \times 3^{2 n+3}-9^{n+1}}$
$ =\frac{3 \times\left(3^2\right)^{n+1}-3^2 \times 3^{2 n}}{3 \times 3^{2 n+3}-\left(3^2\right)^{n+1}}$
$ =\frac{3^{1+2 n+2}-3^{2+2 n}}{3^{1+2 n+3}-3^{2 n+2}}$
$ =\frac{3^{3+2 n}-3^{2+2 n}}{3^{4+2 n}-3^{2 n+2}}$
$ =\frac{3^{2 n}\left(3^3-3^2\right)}{3^{2 n}\left(3^4-3^2\right)}$
$ =\frac{27-9}{81-9}$
$ =\frac{18}{72}$
$ =\frac{1}{4}$
View full question & answer→Question 43 Marks
Evaluate $: \frac{4}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{2}{(243)^{-\frac{1}{5}}}$
Answer$ \frac{4}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{2}{(243)^{-\frac{1}{5}}}$
$ =\frac{4}{\left(6^3\right)^{-\frac{2}{3}}}+\frac{1}{\left(4^4\right)^{-\frac{3}{4}}}+\frac{2}{\left(3^5\right)^{-\frac{1}{5}}}$
$ =\frac{4}{(6)^{-2}}+\frac{1}{(4)^{-3}}+\frac{2}{(3)^{-1}}$
$ =4 \times 6^2+1 \times 4^3+2 \times 3$
$ =4 \times 36+1 \times 64+6$
$ =144+64+6$
$ =214$
View full question & answer→Question 53 Marks
Prove that $: \frac{a+b+c}{a^{-1} b^{-1}+b^{-1} c^{-1}+c^{-1} a^{-1}}=a b c$
Answer$ \text { L.H.S. }=\frac{a+b+c}{a^{-1} b^{-1}+b^{-1} c^{-1}+c^{-1} a^{-1}}$
$ =\frac{a+b+c}{\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}}$
$ =\frac{a+b+c}{\frac{c+a+b}{a b c}}$
$ =\frac{(a+b+c)(a b c)}{a+b+c}$
$ =a b c$
$ =\text { R.H.S. }$
View full question & answer→Question 63 Marks
Evaluate$:\frac{2^n \times 6^{m+1} \times 10^{m-n} \times 15^{m+n-2}}{4^m \times 3^{2 m+n} \times 25^{m-1}}$
Answer$ \frac{2^n \times 6^{m+1} \times 10^{m-n} \times 15^{m+n-2}}{4^m \times 3^{2 m+n} \times 25^{m-1}}$
$ =\frac{2^n \times 6^m \times 6 \times 10^m \times 10^{-n} \times 15^m \times 15^n \times 15^{-2}}{4^m \times\left(3^2\right)^m \times 3^n \times 25^m \times 25^{-1}}$
$ =\frac{\left(2 \times \frac{1}{10} \times 15\right)^n \times(6 \times 10 \times 15)^m \times 6 \times \frac{1}{15^2}}{3^n \times\left(4 \times 3^2 \times 25\right)^m \times \frac{1}{25}}$
$ =\frac{3^n \times 900^m \times \frac{6}{225}}{3^n \times 900^m \times \frac{1}{25}}$
$ =\frac{6}{225} \times \frac{25}{1}$
$ =\frac{6}{9}$
$ =\frac{2}{3}$
View full question & answer→Question 73 Marks
Evaluate $: \left[\left(-\frac{2}{3}\right)^{-2}\right]^3 \times\left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6}$
Answer$ {\left[\left(-\frac{2}{3}\right)^{-2}\right]^3 \times\left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6}}$
$ =\left[\left(-\frac{3}{2}\right)^2\right]^3 \times(3)^4 \times \frac{1}{3} \times \frac{1}{3 \times 2}$
$ =\left(-\frac{3}{2}\right)^6 \times(3)^2 \times \frac{1}{2}$
$ =\frac{3^{6+2}}{2^{6+1}}$
$ =\frac{3^8}{2^7}$
View full question & answer→Question 83 Marks
If $(a^m)^n= a^m.a^n,$ find the value of $: m(n - 1) - (n - 1)$
Answer$(a^m)^n= a^m.a^n$
$\Rightarrow a^{mn} = a^{m + n}$
$\Rightarrow mn = m + n ....(1)$
Now,
$m( n - 1 ) - ( n - 1 )$
$= mn - m - n + 1$
$= m + n - m - n + 1 ....[$ From $(1) ]$
$= 1$
View full question & answer→Question 93 Marks
Evaluate $: 9^{\frac{5}{2}}-3 \times 8^0-\left(\frac{1}{81}\right)^{-\frac{1}{2}}$
Answer$ 9^{\frac{5}{2}}-3 \times 8^0-\left(\frac{1}{81}\right)^{-\frac{1}{2}}$
$ =\left(3^2\right)^{\frac{5}{2}}-3 \times 1-\left(\frac{1}{3^4}\right)^{-\frac{1}{2}}$
$ =3^{2 \times \frac{5}{2}}-3-3^{-4 \times\left(-\frac{1}{2}\right)}$
$ =3^5-3-3^2$
$ =243-3-9$
$ =231$
View full question & answer→Question 103 Marks
Solve$ : 3(2^x+ 1) - 2^{x+2}+ 5 = 0.$
Answer$3(2^x+ 1) - 2^{x+2}+ 5 = 0.$
$\Rightarrow 3 \times 2^x + 3 - 2^x \times 2^2 + 5 = 0$
$\Rightarrow 2^x( 3 - 2^2 ) + 8 = 0$
$\Rightarrow 2^x( 3 - 4 ) = - 8$
$\Rightarrow 2^x x ( - 1 ) = - 8$
$\Rightarrow 2^x = 8$
$\Rightarrow 2^x = 2^3$
$\Rightarrow x = 3$
View full question & answer→Question 113 Marks
If $\mathrm{a}^{\mathrm{x}}=\mathrm{b}^{\mathrm{y}}=\mathrm{c}^{\mathrm{z}}$ and $\mathrm{b}^2=\mathrm{ac},$prove that $\mathrm{y}=\frac{2 a z}{x+z}$
AnswerLet $a x=b y=c z=k$
$\therefore a=k^{\frac{1}{x}} ; b=k^{\frac{1}{y}} ; c=k^{\frac{1}{z}}$
Also, We have $b2 = ac$
$\therefore\left(k^{\frac{1}{y}}\right)^2=\left(k^{\frac{1}{x}}\right) \times\left(k^{\frac{1}{2}}\right)$
$ \Rightarrow k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}$
$ \Rightarrow k^{\frac{2}{y}}=\frac{k^{z+x}}{x z}$
Comparing the powers we have
$\frac{2}{y}=\frac{z+x}{x z}$
$ \Rightarrow y=\frac{2 x z}{z+x}$
View full question & answer→Question 123 Marks
If $a^x= b, b^y= c$ and $c^z = a,$ prove that $: xyz = 1.$
AnswerWe are given that
$a^x = b, b^y = c$ and $c^z = a$
Consider the equation
$a^x = b$
$\Rightarrow a^{xyz} = b^{yz} [$ raising to the power $\text{yz}$ on both sides $]$
$\Rightarrow a^{xyz} = (b^y)^z$
$\Rightarrow a^{xyz} = c^z [ \because b^y = c ]$
$\Rightarrow a^{xyz} = c^z$
$\Rightarrow a^{xyz} = a [ \because c^z = a ]$
$\Rightarrow a^{xyz} = a^1$
$\Rightarrow xyz = 1$
View full question & answer→Question 133 Marks
Prove that$:\frac{x^{a(b-c)}}{x^b(a-c)} \div\left(\frac{x^b}{x^a}\right)^c=1$
AnswerWe need to prove that
$\frac{x^{a(b-c)}}{x^b(a-c)} \div\left(\frac{x^b}{x^a}\right)^c=1$
$ \text { LHS }=$
$ =x^{a(b-c)-b(a-c)} \div \frac{x^{b c}}{x^{a c}}$
$ =x^{a b-a c-a b+b c} \div x^{b c-a c}$
$ =x^{a b-a c-a b+b c-b c+a c}$
$ =x^0$
$ =1$
$ =\text { RHS }$
View full question & answer→Question 143 Marks
Prove that $: \left(\frac{x^a}{x^b}\right)^{a+b-c}\left(\frac{x^b}{x^c}\right)^{b+c-a}\left(\frac{x^c}{x^a}\right)^{c+a-b}$
Answer$ \text { LHS }=\left(\frac{x^a}{x^b}\right)^{a+b-c}\left(\frac{x^b}{x^c}\right)^{b+c-a}\left(\frac{x^c}{x^a}\right)^{c+a-b}$
$ =\left(x^{a-b}\right)^{a+b-c} \times\left(x^{b-c}\right)^{b+c-a} \times\left(x^{c-a}\right)^{c+a-b}$
$ =x^{(a-b)(a+b-c)} \times x^{(b-c)(b+c-a)} \times x^{(c-a)(c+a-b)}$
$ =$
$ x^{a^2+a b-a c-a b-b^2+b c} \times x^{b^2+b c-a b-c d-c^2+a c} \times x^{c^2+a c-b c-a c-a^2+a b}$
$ =x^{a^2-a c-b^2+b c+b^2-a b-c^2+a c+c^2-b c-a^2+a b}$
$ =x^0$
$ =1$
$ =\text { RHS }$
View full question & answer→Question 153 Marks
Solve$ : (\sqrt{3})^{x-3}=(\sqrt[4]{3})^{x+1}$
Answer$ (\sqrt{3})^{x-3}=(\sqrt[4]{3})^{x+1}$
$ \Rightarrow\left(3^{\frac{1}{2}}\right)^{x-3}=\left(3^{\frac{1}{4}}\right)^{x+1}$
$ \Rightarrow 3^{\frac{x-3}{2}}=3^{\frac{x+1}{4}}$
$ \Rightarrow \frac{x-3}{2}=\frac{x+1}{4}$
$ \Rightarrow 4(x-3)=2(x+1)$
$ \Rightarrow 4 x-12=2 x+2$
$ \Rightarrow 4 x-2 x=12+2$
$ \Rightarrow 2 x=14$
$ \Rightarrow x=\frac{14}{2}$
$ \Rightarrow x=7$
View full question & answer→Question 163 Marks
Find $x$, if $: \sqrt{2^{x+3}}=16$
Answer
$\sqrt{2^{x+3}}=16$
$ \left(2^{x+3}\right)^{\frac{1}{2}}=2 \times 2 \times 2 \times 2$
$ \Rightarrow(2)^{\frac{x+3}{2}}=2^4$
We know that if bases are equal, the powers are equal.
$\Rightarrow \frac{x+3}{2}=4$
$ \Rightarrow x+3=8$
$ \Rightarrow x=8-3$
$ \Rightarrow x=5$
View full question & answer→Question 173 Marks
Find $x$, if $: 4^{2 x}=\frac{1}{32}$
Answer$4^{2 x}=\frac{1}{32}$
$ \Rightarrow(2 \times 2)^{2 x}=\frac{1}{2 \times 2 \times 2 \times 2 \times 2}$
$ \Rightarrow\left(2^2\right)^{2 x}=\frac{1}{2^5}$
$ \Rightarrow 2^{2 \times 2 x}=2^{-5}$
$ \Rightarrow 2^{4 x}=2^{-5}$
We know that if bases are equal, the powers are equal
$\Rightarrow 4 x=-5$
$ \Rightarrow x=\frac{-5}{4}$
View full question & answer→Question 183 Marks
Solve for $x$ :$(49)^{x + 4}= 7^2 \times(343)^{x + 1}$
Answer$(49)^{x + 4}= 7^2\times (343)^{x + 1}$
$\Rightarrow ( 7 \times 7 )^{x + 4} = 7^2 ( 7 \times 7 \times 7 )^{( x + 1 )}$
$\Rightarrow ( 7^2 )^{x + 4} = 7^2( 7^3 )^{( x + 1 )}$
$\Rightarrow 7^{( 2x + 8 )} = 7^2 \times 7^{3x + 3}$
$\Rightarrow 7^{( 2x + 8 )} = 7^{3x + 3 + 2}$
$\Rightarrow 7^{( 2x + 8 )} = 7^{3x + 5}$
We know that if bases are equal, the powers are equal
$\Rightarrow 2x + 8 = 3x + 5$
$\Rightarrow 3x - 2x = 8 - 5$
$\Rightarrow x = 3$
View full question & answer→Question 193 Marks
Solve for $\mathbf{x}: 4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{-x}$
Answer$ 4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{-x}$
$ \Rightarrow\left(2^2\right)^{x-1} \times\left(\frac{1}{2}\right)^{3-2 x}=\left(\frac{1}{2^3}\right)^{-x}$
$ \Rightarrow 2^{2 x-2} \times 2^{-(3-2 x)}=\left(2^{-3}\right)^{-x}$
$ \Rightarrow 2^{2 x-2-3+2 x}=2^{3 x}$
$ \Rightarrow 2^{4 x-5}=2^{3 x}$
$ \Rightarrow 4 x-5=3 x$
$ \Rightarrow 4 x-3 x=5$
$ \Rightarrow x=5$
View full question & answer→Question 203 Marks
If $5^{x + 1}= 25^{x- 2},$ find the value of $3^{x - 3}\times 2^{3 - x}.$
Answer$5^{x+1}=25^{x-2}$
$ \Rightarrow 5^{x+1}=\left(5^2\right)^{x-2}$
$ \Rightarrow 5^{x+1}=5^{2 x-4}$
If bases are equal, powers are also equal.
$\Rightarrow x+1=2 x-4$
$ \Rightarrow 2 x-x=4+1$
$ \Rightarrow x=5$
$\therefore 3^{x-3} \times 2^{3-x}$
$ =3^{5-3} \times 2^{3-5}$
$ =3^2 \times 2^{-2}$
$ =9 \times \frac{1}{4}$
$=\frac{9}{4}$
View full question & answer→Question 213 Marks
Solve for $x : 2^{5x-1}= 42^{3x + 1}$
Answer$2^{5 x-1}=4 \times 2^{3 x+1}$
$ \Rightarrow 2^{5 x-1}=2^2 \times 2^{3 x+1}$
$ \Rightarrow 2^{5 x-1}=2^{2+3 x+1}$
$ \Rightarrow 2^{5 x-1}=2^{3 x+3}$
We know that if bases are equal, the powers are equal
$\Rightarrow 5 x-1=3 x+3$
$ \Rightarrow 5 x-3 x=3+1$
$ \Rightarrow 2 x=4$
$ \Rightarrow x=\frac{4}{2}$
$ \Rightarrow x=2 .$
View full question & answer→Question 223 Marks
If $5^{-P}=4^{-q}=20^r,$ show that $: \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0$
Answer$ \text { Let } 5^{-\mathrm{P}}=4^{-\mathrm{q}}=20^{\mathrm{r}}=\mathrm{k}$
$ 5^{-\mathrm{P}}=\mathrm{k}$
$\Rightarrow 5=k^{-\frac{1}{p}} \quad\left[\because a^p=b^q \Rightarrow a=b^{\frac{q}{p}}\right]$
$ 4^{-\mathrm{q}}=\mathrm{k}$
$\Rightarrow 4=k^{-\frac{1}{q}}\left[\because a^p=b^q \Rightarrow a=b^{\frac{q}{p}}\right]$
$ 20^{-\mathrm{r}}=\mathrm{k}$
$\Rightarrow 20=k^{\frac{1}{r}} \quad\left[\because a^p=b^q \Rightarrow a=b^{\frac{q}{p}}\right]$
$5 \times 4=20$
$ \Rightarrow k^{-\frac{1}{p}} \times k^{-\frac{1}{q}}=k^{\frac{1}{r}}$
$ \Rightarrow k^{-\frac{1}{p}-\frac{1}{q}}=k^{\frac{1}{r}}$
$ \Rightarrow k^0=k^{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}}$
If bases are equal, powers are also equal.
$\Rightarrow \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0$
View full question & answer→Question 233 Marks
Show that$:\left(\frac{a^m}{a^{-n}}\right)^{m-n} \times\left(\frac{a^n}{a^{-l}}\right)^{n-l} \times\left(\frac{a^l}{a^{-m}}\right)^{l-m}=1$
Answer$ \left(\frac{a^m}{a^{-n}}\right)^{m-n} \times\left(\frac{a^n}{a^{-l}}\right)^{n-l} \times\left(\frac{a^l}{a^{-m}}\right)^{l-m}=1$
$ =\left(a^m \times a^n\right)^{m-n} \times\left(a^n \times a^l\right)^{n-l} \times\left(a^l \times a^m\right)^{l-m}$
$ =\left(a^{m+n}\right)^{m-n} \times\left(a^{n+l}\right)^{n-l} \times\left(a^{l+m}\right)^{l-m}$
$ =a^{m^2-n^2} \times a^{n^2-l^2} \times a^{l^2-m^2}$
$ =a^{m^2-n^2+n^2-l^2+l^2-m^2}$
$ =a^0$
$ =1$
View full question & answer→Question 243 Marks
Simplify $ :\frac{8^3 a \times 2^5 \times 2^{2 a}}{4 \times 2^{11 a} \times 2^{-2 a}}$
Answer$ \frac{8^3 a \times 2^5 \times 2^{2 a}}{4 \times 2^{11 a} \times 2^{-2 a}}$
$ =\frac{\left(2^3\right)^{3 a} \times 2^5 \times 2^{2 a}}{2^2 \times 2^{11 a} \times 2^{-2 a}}$
$ =\frac{2^{3 \times 3 a} \times 2^5 \times 2^{2 a}}{2^2 \times 2^{11 a} \times 2^{-2 a}}$
$ =\frac{2^{9 a} \times 2^5 \times 2^{2 a}}{2^2 \times 2^{11 a} \times 2^{-2 a}}$
$ =2^{9 a+5+2 a-2-11 a+2 a}$
$ =2^{2 a+3}$
View full question & answer→Question 253 Marks
Simplify the following and express with positive index $:(32)^{-\frac{2}{5}} \div(125)^{-\frac{2}{5}}$
Answer$ (32)^{-\frac{2}{5}} \div(125)^{-\frac{2}{5}}$
$ =\left[\frac{(32)^{-\frac{2}{5}}}{(125)^{-\frac{2}{3}}}\right]$
$ =\frac{(125)^{\frac{2}{3}}}{(32)^{\frac{2}{5}}}$
$ =\frac{(5 \times 5 \times 5)^{\frac{2}{3}}}{(2 \times 2 \times 2 \times 2 \times 2)^{\frac{2}{5}}}$
$ =\frac{\left(5^3\right)^{\frac{2}{3}}}{\left(2^5\right)^{\frac{2}{5}}}$
$ =\frac{5^2}{2^2}$
$ =\frac{25}{4}$
$=\left(\frac{5}{2}\right)^2$
View full question & answer→Question 263 Marks
Simplify the following and express with positive index $:\left(\frac{27^{-3}}{9^{-3}}\right)^{\frac{1}{5}}$
Answer$ \left(\frac{27^{-3}}{9^{-3}}\right)^{\frac{1}{5}}$
$ =\left(\frac{9^3}{27^3}\right)^{\frac{1}{5}}$
$ =\left[\frac{\left(3^2\right)^3}{\left(3^3\right)^3}\right]^{\frac{1}{5}}$
$ =\left[\left(\frac{3^2}{3^3}\right)^3\right]^{\frac{1}{5}}$
$ =\left[\left(\frac{1}{3}\right)^3\right]^{\frac{1}{5}}$
$ =\left(\frac{1}{3}\right)^{3 \times \frac{1}{5}}$
$ =\frac{1}{(3)^{\frac{3}{5}}}$
View full question & answer→Question 273 Marks
Simplify $:\left(8 x^3 \div 125 y^3\right)^{\frac{2}{3}}$
Answer$ \left(8 x^3 \div 125 y^3\right)^{\frac{2}{3}}$
$ =\left(\frac{2 x \times 2 x \times 2 x}{5 y \times 5 y \times 5 y}\right)^{\frac{2}{3}}$
$ =\left[\left(\frac{2 x}{5 y}\right)^3\right]^{\frac{2}{3}}$
$ =\left(\frac{2 x}{5 y}\right)^{3 \times \frac{2}{3}}$
$ =\left(\frac{2 x}{5 y}\right)^2$
$ =\left[\frac{2 x}{5 y} \times \frac{2 x}{5 y}\right]$
$ =\frac{4 x^2}{25 y^2}$
View full question & answer→Question 283 Marks
Evaluate$:7^0 \times(25)^{-\frac{3}{2}}-5^{-3}$
Answer$ 7^0 \times(25)^{-\frac{3}{2}}-5^{-3}$
$ =7^0 \times(5 \times 5)^{-\frac{3}{2}}-5^{-3}$
$ =7^0 \times\left(5^2\right)^{-\frac{3}{2}}-\frac{1}{5^3}$
$ =7^0 \times\left[(5)^{2 \times\left(-\frac{3}{2}\right)}\right]-\frac{1}{5^3}$
$ =7^0 \times 5^{-3}-\frac{1}{5^3}$
$ =1 \times 5^{-3}-\frac{1}{5^3}$
$ =\frac{1}{5^3}-\frac{1}{5^3}$
$ =\frac{1-1}{5 \times 5 \times 5}$
$ =\frac{0}{125}$
$ =0$
View full question & answer→Question 293 Marks
Evaluate$:\left(\frac{27}{125}\right)^{\frac{2}{3}} \times\left(\frac{9}{25}\right)^{-\frac{3}{2}}$
Answer$ \left(\frac{27}{125}\right)^{\frac{2}{3}} \times\left(\frac{9}{25}\right)^{-\frac{3}{2}}$
$ =\left(\frac{3 \times 3 \times 3}{5 \times 5 \times 5}\right)^{\frac{2}{3}} \times\left(\frac{3 \times 3}{5 \times 5}\right)^{-\frac{3}{2}}$
$ =\left[\left(\frac{3}{5}\right)^3\right]^{\frac{2}{3}} \times\left[\left(\frac{3}{5}\right)^2\right]^{-\frac{3}{2}}$
$ =\left(\frac{3}{5}\right)^{3 \times \frac{2}{3}} \times\left(\frac{3}{5}\right)^{2 \times-\frac{3}{2}}$
$ =\left(\frac{3}{5}\right)^2 \times\left(\frac{3}{5}\right)^{-3}$
$ =\left(\frac{3}{5}\right)^{2-3}$
$ =\left(\frac{3}{5}\right)^{-1}$
$ =\frac{1}{\frac{3}{5}}$
$ =\frac{5}{3}$
View full question & answer→Question 303 Marks
Evaluate:$5^{-4} \times(125)^{\frac{5}{3}} \div(25)^{-\frac{1}{2}}$
Answer$ 5^{-4} \times(125)^{\frac{5}{3}} \div(25)^{-\frac{1}{2}}$
$ =5^{-4} \times(5 \times 5 \times 5)^{\frac{5}{3}} \div(5 \times 5)^{-\frac{1}{2}}$
$ =5^{-4} \times\left(5^3\right)^{\frac{5}{3}} \div\left(5^2\right)^{-\frac{1}{2}}$
$ =5^{-4} \times\left[(5)^{3 \times \frac{5}{3}}\right] \div\left[(5)^{2 \times-\frac{1}{2}}\right]$
$ =\frac{5^{-4} \times 5^5}{5^{-1}}$
$ =\frac{5^{5-4}}{5^{-1}}$
$ =\frac{5^1}{5^{-1}}$
$ =5^{1-(-1)}$
$ =5^2$
$ =5 \times 5$
$ =25$
View full question & answer→Question 313 Marks
Evaluate$:3^3 \times(243)^{-\frac{2}{3}} \times 9^{-\frac{1}{3}}$
Answer$ 3^3 \times(243)^{-\frac{2}{3}} \times 9^{\frac{1}{3}}$
$ =3^3 \times(3 \times 3 \times 3 \times 3 \times 3)^{-\frac{2}{3}} \times(3 \times 3)^{-\frac{1}{3}}$
$ =3^3 \times\left(3^5\right)^{-\frac{2}{3}} \times\left(3^2\right)^{\frac{1}{3}}$
$ =3^3 \times 3^{-\frac{10}{3}} \times 3^{-\frac{2}{3}} \ldots\left[\left(a^m\right)^n=a^{m n}\right]$
$ =3^{3-\frac{10}{3}-\frac{2}{3}} \quad\left[a^m \times a^n \times a^o=a^{m+n+o}\right]$
$ =3^{\frac{9-10-2}{3}}$
$ =3^{\frac{9-12}{3}}$
$ =3^{-\frac{3}{3}}$
$ =3^{-1}$
$ =\frac{1}{3}$
View full question & answer→Question 323 Marks
Simplify $:\left(\frac{x^a}{x^{-b}}\right)^{a^2-a b+b^2} \times\left(\frac{x^b}{x^{-c}}\right)^{b^2-b c+c^2} \times\left(\frac{x^c}{x^{-a}}\right)^{c^2-c a+a^2}$
Answer$ \left(\frac{x^a}{x^{-b}}\right)^{a^2-a b+b^2} \times\left(\frac{x^b}{x^{-c}}\right)^{b^2-b c+c^2} \times\left(\frac{x^c}{x^{-a}}\right)^{c^2-c a+a^2}$
$ =\left(x^{a+b}\right)^{a^2-a b+b^2} \times\left(x^{b+c}\right)^{b^2-b c+c^2} \times\left(x^{c+a}\right)^{c^2-c a+a^2}$
$ =x^{a^3+b^3} \times x^{b^3+c^3} \times\left(x^{c^3+a^3}\right)$
$ =x^{a^3+b^3+b^3+c^3+c^3+a^3}$
$ =x^{\left(a^3+b^3+b^3+c^3+c^3+a^3\right)}$
$ =x^{2\left(a^3+b^3+c^3\right)}$
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Simplify $:\left(\frac{x^a}{x^b}\right)^{a^2+a b+b^2} \times\left(\frac{x^b}{x^c}\right)^{b^2+b c+c^2} \times\left(\frac{x^c}{x^a}\right)^{c^2+c a+a^2}$
Answer$ \left(\frac{x^a}{x^b}\right)^{a^2+a b+b^2} \times\left(\frac{x^b}{x^c}\right)^{b^2+b c+c^2} \times\left(\frac{x^c}{x^a}\right)^{c^2+c a+a^2}$
$ =\left(x^{a-b}\right)^{a^2+a b+b^2} \times\left(x^{b-c}\right)^{b^2+b c+c^2} \times\left(x^{c-a}\right)^{c^2+c a+a^2}$
$ =x^{a^3-b^3} \times x^{b^3-c^3} \times x^{c^3-a^3}$
$ =x^{a^3-b^3+b^3-c^3+c^3-a^3}$
$ =x^0$
$ =1$
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