Question

Find the value of $(243{)}^{-\frac{4}{5}}$.

• A. $81$
• B. $\frac{1}{81}$
  
• C. $27$
• D. $\frac{1}{27}$
  

Answer 1

The correct option is B $\frac{1}{81}$


$\underset{–}{\text{Need to find:}}$ $(243{)}^{-\frac{4}{5}}$.

To eliminate the negative sign from the exponent, we would take the fraction's (base) reciprocal.

$(243{)}^{-\frac{4}{5}}=(\frac{1}{243}{)}^{\frac{4}{5}}$
$[\because (\frac{a}{b}{)}^{-n}=(\frac{b}{a}{)}^n]$

Now,
$(\frac{1}{243}{)}^{\frac{4}{5}}$

$=(\frac{1^5}{3^5}{)}^{\frac{4}{5}}$
$(\because 1^5=1\times 1\times 1\times 1\times 1=1\text{and}243=3\times 3\times 3\times 3\times 3=3^5)$

$=[(\frac{1}{3}{)}^5{]}^{\frac{4}{5}}$
$(\because \frac{a^n}{b^n}=(\frac{a}{b}{)}^n)$

$=(\frac{1}{3}{)}^{5\times \frac{4}{5}}$
$(\because (a^m{)}^n=a^{m\times n})$

$=(\frac{1}{3}{)}^4$

$=\frac{1^4}{3^4}$
$(\because (\frac{a}{b}{)}^n=\frac{a^n}{b^n})$

$=\frac{1\times 1\times 1\times 1}{3\times 3\times 3\times 3}$

$\Rightarrow \underset{–}{(\frac{1}{243}{)}^{\frac{4}{5}}=(243{)}^{-\frac{4}{5}}=\frac{1}{81}}$

Hence, the value of $(243{)}^{-\frac{4}{5}}$ is $\frac{1}{81}.$

$\therefore$ Option (b.) is the correct answer.