# How do you solve 4^x = 7^(x-4)?

Dec 4, 2015

$x \cong - 6.7745$

#### Explanation:

Given the exponential equation ${4}^{x} = {7}^{x - 4}$

To solve exponential equation we can use logarithm.

Step 1: Take log of both side

$\log {4}^{x} = \log {7}^{x - 4}$

Using the power rule of logarithm

$x \log 4 = \left(x - 4\right) \log 7$

Then distribute

$x \log 4 = x \log 7 - 4 \log 7$

Then bring all the "x" on one side

$x \log 4 - x \log 7 = - 4 \log 7$

Factor out the greatest common factor

$x \left(\log 4 - \log 7\right) = - 4 \log 7$

Isolate "x"

$x = \frac{- 4 \log 7}{\log 4 - \log 7}$

$x \cong - 6.7745$