Question

If ${243}^{\frac{2}{y}}=9$, find the value of $y(y\ne 0)$.

• A. $y=4$
• B. $y=5$
• C. $y=3$
• D. $y=9$

Answer 1

The correct option is B $y=5$

We need to find the value of $y$ in the equation ${243}^{\frac{2}{y}}=9$.

$243$ can be written as $243=81\times 3243=9\times 9\times 3243=9^2\times 3$
Putting this in the original equation:
${243}^{\frac{2}{y}}=9$

$\Rightarrow (9^2\times 3{)}^{\frac{2}{y}}=9$

$\Rightarrow ((9^2{)}^2\times (3{)}^2{)}^{\frac{1}{y}}=9$

$\Rightarrow (9^4\times 9{)}^{\frac{1}{y}}=9$

$\Rightarrow (9^{(4+1)}{)}^{\frac{1}{y}}=9$

$[\because a^m\times a^n=a^{(m+n)}]$

$\Rightarrow (9^5{)}^{\frac{1}{y}}=9$

$\Rightarrow 9^{\frac{5}{y}}=9$

$\Rightarrow 9^{\frac{5}{y}}=9^1$

Exponents with equal bases are equal only when powers are equal

$\frac{5}{y}=1$

As $y\ne 0$, we can multiply both sides of the above equation by $y$.

$y\times \frac{5}{y}=y\times 1$

$\Rightarrow 5=y$

$\Rightarrow y=5$

Thus, the value of $y$ in the equation ${243}^{\frac{2}{y}}=9$ is $5$.

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