# How do you solve 8^(2x-5)=5^(x+1)?

Dec 10, 2016

$x = \frac{5 \ln \left(8\right) + \ln \left(5\right)}{2 \ln \left(8\right) - \ln \left(5\right)} \approx 4.7095$

#### Explanation:

Using the property of logarithms that $\log \left({a}^{x}\right) = x \log \left(a\right)$, we have

${8}^{2 x - 5} = {5}^{x + 1}$

$\implies \ln \left({8}^{2 x - 5}\right) = \ln \left({5}^{x + 1}\right)$

$\implies \left(2 x - 5\right) \ln \left(8\right) = \left(x + 1\right) \ln \left(5\right)$

$\implies 2 \ln \left(8\right) x - 5 \ln \left(8\right) = \ln \left(5\right) x + \ln \left(5\right)$

$\implies 2 \ln \left(8\right) x - \ln \left(5\right) x = 5 \ln \left(8\right) + \ln \left(5\right)$

$\implies \left(2 \ln \left(8\right) - \ln \left(5\right)\right) x = 5 \ln \left(8\right) + \ln \left(5\right)$

$\therefore x = \frac{5 \ln \left(8\right) + \ln \left(5\right)}{2 \ln \left(8\right) - \ln \left(5\right)} \approx 4.7095$