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

Apr 2, 2016

For this problem, we must use the rule ${a}^{n} = {b}^{m} \to \log {a}^{n} = \log {b}^{m}$

#### Explanation:

${4}^{2 x + 3} = 1$

$\log {4}^{2 x + 3} = \log 1$

Use the property $\log {n}^{a} = a \log n$

$\left(2 x + 3\right) \log 4 = 0$

$2 x + 3 = \frac{0}{\log} 4$

$2 x + 3 = 0$

$2 x = - 3$

$x = - \frac{3}{2}$

Checking we find that the solution works, since ${a}^{0} = 1 \to {4}^{0} = 1$

Practice exercises:

Solve for x. Express your answer in terms of $\log x$ (exact values)

a) ${2}^{3 x + 1} = 5$

b) ${3}^{2 x - 1} = {4}^{3 x - 7}$

Challenge Problem

Solve ${2}^{3 x} \times 5 = {3}^{4 x - 1}$

Good luck!