Question

What is the solution to the equation below?
3log4x=log432+log42
X=-8
X=-4
X=4
X=8

Answer 1

Answer:

$\large\boxed{x=4}$

Step-by-step explanation:

$3\log_4x=\log_432+\log_42\qquad\text{use}\ \log_ab^c=c\log_ab\ \text{and}\ \log_ab+\log_ac=\log_a(bc)\\\\\log_4x^3=\log_4(32\cdot2)\\\\\log_4x^3=\log_464\iff x^3=64\to x=\sqrt[3]{64}\\\\x=4$