# How do you solve 4^(2x+3) = 1?

Feb 24, 2016

Convert to logarithmic form.

#### Explanation:

$\log \left({4}^{2 x + 3}\right) = 1$

Simplify using the rule $\log {n}^{a} = a \log n$

$\left(2 x + 3\right) \log 4 = 1$

You can now distribute. Don't forget: you can't combine logarithmic and non logarithmic expressions.

$2 x \log 4 + 3 \log 4 = 1$

$2 x \log 4 = 1 - 3 \log 4$

Factor out the x.

$x \left(2 \log 4\right) = 1 - \log {4}^{3}$

$x = \frac{1 - \log 64}{\log} 16$

Ask your teacher exactly what he/she wants. They may want you to keep your answer in exact form, or for you to round it off to a certain number of decimals, so just make sure to communicate.

Practice exercises:

1. Solve for x.

a) ${2}^{3 x - 7} = 5$

b). ${3}^{2 x + 3} = {4}^{x - 5}$

c) ${2}^{4 x - 9} = 3 \times {3}^{x + 6}$

Good luck!