[tex]\left(\dfrac{9a^{-4}}{25}\right)^p=\dfrac3{5a^2}[/tex]
[tex]\left(\dfrac9{25a^4}\right)^p=\dfrac3{5a^2}[/tex]
One way to find [tex]p[/tex] is to notice that the part on the left hand side within the parentheses can be written as a square:
[tex]\dfrac9{25a^4}=\dfrac{3^2}{5^2(a^2)^2}=\left(\dfrac3{5a^2}\right)^2[/tex]
So you could write
[tex]\left(\left(\dfrac3{5a^2}\right)^2\right)^p=\left(\dfrac3{5a^2}\right)^{2p}=\dfrac3{5a^2}[/tex]
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 [tex]2p=1[/tex], or [tex]p=\dfrac12[/tex].
