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

Feb 11, 2016

You must first convert to logarithmic form.

#### Explanation:

$\log {5}^{x - 1} = \log {3}^{x}$

$\left(x - 1\right) \log 5 = x \log 3$

Distribute the parentheses:

$x \log 5 - \log 5 = x \log 3$

Put all x terms to the left side of the equation:

$x \log 5 - x \log 3 = \log 5$

Factor out the x:

$x \left(\log 5 - \log 3\right) = \log 5$

Use the quotient rule:

$x \left(\log \left(\frac{5}{3}\right)\right) = \log 5$

x = log5/(log(5/3)

$x = {\log}_{\frac{5}{3}} 5$

Practice exercises:

1. Solve for x. Leave in logarithmic form.

a) ${2}^{x - 2} = {3}^{2 x + 4}$

b) ${3}^{3 x} = 4 \times {5}^{x - 6}$

Good luck!