# How do you solve 7^x=4^(x+2)?

Mar 16, 2016

You must convert to logarithmic form.

#### Explanation:

${7}^{x} = {4}^{x + 2}$

$\log {7}^{x} = \log {4}^{x + 2}$

You must now use the rule $\log {a}^{x} = x \log a$ to simplify further.

$x \log 7 = \left(x + 2\right) \log 4$

$x \log 7 = x \log 4 + 2 \log 4$

$x \log 7 - x \log 4 = \log {4}^{2}$

$x \left(\log 7 - \log 4\right) = \log 16$

You can now simplify even further by using the rule ${\log}_{a} n - {\log}_{a} m = {\log}_{a} \left(\frac{n}{m}\right)$

$x = \log \frac{16}{\log \left(\frac{7}{4}\right)}$

This is as simplified as it gets. Make sure to ask your teacher if they prefer your answer to be in logarithmic form (as shown above) or as an approximate value (rounded to an x number of decimal places).

Hopefully this helps!