# How do you solve 4^x=6^(x+2)?

Mar 24, 2016

In ${4}^{x} = {6}^{x + 2}$, taking log of both the sides, we get

$x \times \log 4 = \left(x + 2\right) \times \log 6$ or

$x \left(\log 4 - \log 6\right) = 2 \log 6$ or

$x = \frac{2 \log 6}{\log 4 - \log 6}$

Now we can simplify using logarithmic tables.

As $\log 4 = 0.6021$ and $\log 6 = 0.7782$

$x = \frac{2 \times 0.7782}{0.6021 - 0.7782}$ or

$x = \frac{1.5564}{-} 0.1761 = - 8.838$