Question

Solve for 'p'

Answer 1

$\left(\dfrac{9a^{-4}}{25}\right)^p=\dfrac3{5a^2}$
$\left(\dfrac9{25a^4}\right)^p=\dfrac3{5a^2}$

One way to find $p$ is to notice that the part on the left hand side within the parentheses can be written as a square:

$\dfrac9{25a^4}=\dfrac{3^2}{5^2(a^2)^2}=\left(\dfrac3{5a^2}\right)^2$

So you could write

$\left(\left(\dfrac3{5a^2}\right)^2\right)^p=\left(\dfrac3{5a^2}\right)^{2p}=\dfrac3{5a^2}$

Since these two expressions are equal, and the bases are the same, you know the exponents must be equal, with the exponent on the right hand side being 1. So $2p=1$, or $p=\dfrac12$.