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

Mar 28, 2016

You must use logarithms to solve.

#### Explanation:

If ${a}^{n} = b$, then $\log {a}^{n} = \log b$.

$\log {4}^{2 x} = \log 6$

Use the log rule $\log {a}^{n} = n \log a$ to simplify further.

$2 x \log 4 = \log 6$

$x \left(\log {4}^{2}\right) = \log 6$

$x = \frac{\log 6}{\log 16}$

$x = 0.65$

Make sure to ask your teacher if they want your answer in exact form ($x = \frac{\log 6}{\log 16}$) or rounded off to an x number of decimal points. I know that my teacher asks for two decimals after the point ($x = 0.65$), but some ask for 1, 3 or 0.

Practice exercises:

1. Solve for x:

a) 2^(3x - 2) = 17

b) 3^(4x + 1) = 6^(3x - 3)

Challenge problem:

Solve for x in ${2}^{2 x - 3} + 5 = {3}^{3 x}$

Hint: think of the log addition rule.

Good luck!