Answer to Question #163631 in Algebra for Samantha Bluhm

Question

Log_4x2^a=3 How do I solve this?

Answer 1

Solution.

$$log_{4x}{2^a}=3$$

Find the domain of the function:

$4x>0, 4x\neq 1\implies \newline x>0,x\neq \frac{1}{4},$ and

$2^a>0\implies a \in R.$

Solve the equation:

$$(4x)^3=2^a,$$ $$x^3=\frac{2^a}{4^3},$$ $$x^3=2^{a-6},$$ $$x=\sqrt[3]{2^{a-6}}.$$

$\sqrt[3]{2^{a-6}} \neq \frac{1}{4},$

$\frac{a-6}{3}\neq -2,$

$a\neq 0.$

Answer. $x=\sqrt[3]{2^{a-6}}, \text{if } a\neq 0.$