Question

If ${27}^k=\frac{9}{3^k},$ then find the value of $\frac{1}{k^2}$.

Answer 1

Since, we have

${27}^k=\frac{9}{3^k}$

$3^{3k}=\frac{3^2}{3^k}$

$3^{3k}=3^{2-k}$

$\therefore 3k=2-k$

Since the base is same, therefore we have
$k=\frac{1}{2}$

$\therefore \frac{1}{k^2}=4$