Question

What is the solution to log2(2x^3-8)-2log2x=log2x?

Answer 1

Answer:

x = 2

Step-by-step explanation:

$\\\\\text{Domain:}\\\\2x^3-8>0\ \wedge\ x>0\\2x^2>8\to x^3>4\to x>\sqrt[3]4\\\\x\in(\sqrt[3]{4},\ \infty)\\\\\log_2(2x^3-8)-2\log_2x=\log_2x\qquad\text{add}\ 2\log_2x\ \text{to both sides}\\\\\log_2(2x^3-8)=3\log_2x\qquad\text{use}\ n\log_ab=\log_ab^n\\\\\log_2(2x^3-8)=\log_2x^3\iff2x^3-8=x^3\qquad\text{subtract}\ 2x^3\ \text{from both sides}\\\\-8=-x^3\qquad\text{change the signs}\\\\8=x^3\to x=\sqrt[3]8\\\\x=2\in D$