# How do you solve 2 times 3^x = 7 times 5^x?

Feb 28, 2016

I have taken you up to the final calculation!

#### Explanation:

Given: $\text{ } 2 \times {3}^{x} = 7 \times {5}^{x}$

Collecting like terms

$\frac{2}{7} = {5}^{x} / {3}^{x}$

Taking logs of both sides ( have chosen to use Natural logs)

$\textcolor{b r o w n}{\text{Note that when taking logs, division of the source values becomes subtraction of the logs.}}$
$\textcolor{b r o w n}{\text{Also "log(b^x) " can be written as } x \log \left(b\right)}$

$\ln \left(2\right) - \ln \left(7\right) = \ln \left({5}^{x}\right) - \ln \left({3}^{x}\right)$

This may be can be written as:

$\ln \left(2\right) - \ln \left(7\right) = x \ln \left(5\right) - x \ln \left(3\right)$

$\ln \left(2\right) - \ln \left(7\right) = x \left(\ln \left(5\right) - \ln \left(3\right)\right)$

$x = \frac{\ln \left(2\right) - \ln \left(7\right)}{\ln \left(5\right) - \ln \left(3\right)}$

I will let you work this out!