# How do you solve 2^(x+3)=5^(3x-1)?

Sep 19, 2016

$x = \ln \frac{40}{\ln} 62.5 \cong 0.892$

#### Explanation:

Take the natural logarithm of both sides:

$\ln \left({2}^{x + 3}\right) = \ln \left({5}^{3 x - 1}\right)$

Apply the rule $\ln \left({a}^{n}\right) = n \ln a$:

$\left(x + 3\right) \ln 2 = \left(3 x - 1\right) \ln 5$

$x \ln 2 + 3 \ln 2 = 3 x \ln 5 - \ln 5$

Isolate the x's to one side:

$3 \ln 2 + \ln 5 = 3 x \ln 5 - x \ln 2$

$3 \ln 2 + \ln 5 = x \left(3 \ln 5 - \ln 2\right)$

Re-condense using the rule $n \ln a = \ln \left({a}^{n}\right)$.

$\ln 8 + \ln 5 = x \left(\ln 125 - \ln 2\right)$

Simplify both sides using the rule $\ln a + \ln b = \ln \left(a \times b\right)$ and $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$

$\ln \left(8 \times 5\right) = x \left(\ln \left(\frac{125}{2}\right)\right)$

$\ln \left(40\right) = x \left(\ln \left(\frac{125}{2}\right)\right)$

$x = \ln \frac{40}{\ln} 62.5$

If you want an approximation, you will get $x \cong 0.892$.

Hopefully this helps!