# How do I use logarithms to solve for x if 7^(2x-1)=316?

Sep 11, 2014

Please take a look at this video that gives step-by-step instructions to solve a very similar problem.

${7}^{2 x - 1} = 316$

Take the natural log of both sides of the equation

$\ln \left({7}^{2 x - 1}\right) = \ln \left(316\right)$

Use the properties of logarithms to move the exponent

$\left(2 x - 1\right) \cdot \ln \left(7\right) = \ln \left(316\right)$

Use basic algebra skills to solve

$\frac{\left(2 x - 1\right) \cdot \ln \left(7\right)}{\ln} \left(7\right) = \ln \frac{316}{\ln} \left(7\right)$

$2 x - 1 = \ln \frac{316}{\ln} \left(7\right)$

$2 x = \ln \frac{316}{\ln} \left(7\right) + 1$

$\frac{2 x}{2} = \frac{\ln \frac{316}{\ln} \left(7\right) + 1}{2}$

$x = \frac{\ln \frac{316}{\ln} \left(7\right) + 1}{2}$

$x = 1.978933191$