# How do you solve 5^(5y-2)=2^(2y+1)?

Oct 16, 2016

${5}^{5 y - 2} = {2}^{2 y + 1}$

$\ln \left({5}^{5 y - 2}\right) = \ln \left({2}^{2 y + 1}\right)$

$\left(5 y - 2\right) \ln 5 = \left(2 y + 1\right) \ln 2$

$5 y \ln 5 - 2 \ln 5 = 2 y \ln 2 + \ln 2$

$5 y \ln 5 - 2 y \ln 2 = \ln 2 + 2 \ln 5$

$y \left(5 \ln 5 - 2 \ln 2\right) = \ln 2 + 2 \ln 5$

Apply the rules $a \ln n = \ln {n}^{a}$, $\ln a - l a b = \ln \left(\frac{a}{b}\right)$ and $\ln a + \ln b = \ln \left(a \times b\right)$ to simplify.

$y \left(\ln 781.25\right) = \ln 50$

$y = \ln \frac{50}{\ln} 781.25$

$y = 0.59$

Hopefully this helps!