Kinda lost on this problem.
Would appreciate some help. Does this have to do with log properties?
-Auburn
No need for logarithms, just convert everything to the same base.
$\displaystyle \displaystyle \frac{25^{3x^2}}{5^{7x}} = \frac{1}{25}$
$\displaystyle \displaystyle \frac{(5^2)^{3x^2}}{5^{7x}} = \frac{1}{5^2}$
$\displaystyle \displaystyle \frac{5^{6x^2}}{5^{7x}} = 5^{-2}$
$\displaystyle \displaystyle 5^{6x^2 - 7x} = 5^{-2}$
$\displaystyle \displaystyle 6x^2 - 7x = -2$
$\displaystyle \displaystyle 6x^2 - 7x + 2 = 0$
$\displaystyle \displaystyle 6x^2 - 3x - 4x + 2 = 0$
$\displaystyle \displaystyle 3x(2x - 1) - 2(2x - 1) = 0$
$\displaystyle \displaystyle (2x-1)(3x-2) = 0$
$\displaystyle \displaystyle 2x-1 = 0$ or $\displaystyle \displaystyle 3x-2 = 0$
$\displaystyle \displaystyle x = \frac{1}{2}$ or $\displaystyle \displaystyle x = \frac{2}{3}$.