Question

Find the value of $((\frac{81}{256}{)}^{\frac{3}{4}}{)}^{-\frac{5}{3}}.$

• A. $\frac{243}{1024}$
  
• B. $-\frac{243}{1024}$
  
• C. $-\frac{1024}{243}$
  
• D. $\frac{1024}{243}$
  

Answer 1

The correct option is D $\frac{1024}{243}$


First we need to simplify the exponents.

$((\frac{81}{256}{)}^{\frac{3}{4}}{)}^{-\frac{5}{3}}=(\frac{81}{256}{)}^{\frac{3}{4}\times (-\frac{5}{3})}$
$[\because (a^m{)}^n=a^{m\times n}]$

$=(\frac{3^4}{4^4}{)}^{-\frac{5}{4}}$ $(\because 81=3\times 3\times 3\times 3=3^4\text{and}256=4\times 4\times 4\times 4=4^4)$

$=[(\frac{3}{4}{)}^4{]}^{-\frac{5}{4}}$ $(\because \frac{a^n}{b^n}=(\frac{a}{b}{)}^n)$

$=(\frac{3}{4}{)}^{4\times (-\frac{5}{4})}$ $[\because (a^m{)}^n=a^{m\times n}]$

$=(\frac{3}{4}{)}^{-5}$

$=(\frac{4}{3}{)}^5$ $[\because (\frac{a}{b}{)}^{-n}=(\frac{b}{a}{)}^n]$

$=\frac{4^5}{3^5}$ $[\because (\frac{a}{b}{)}^n=\frac{a^n}{b^n}]$

$=\frac{4\times 4\times 4\times 4\times 4}{3\times 3\times 3\times 3\times 3}$

$\Rightarrow \underset{–}{((\frac{81}{256}{)}^{\frac{3}{4}}{)}^{-\frac{5}{3}}=\frac{1024}{243}}$

Hence, the value of $((\frac{81}{256}{)}^{\frac{3}{4}}{)}^{-\frac{5}{3}}$ is equal to $\frac{1024}{243}$.

Therefore, option (d.) is the correct answer.