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8_murik_8 [283]
3 years ago
6

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

Mathematics
1 answer:
Alexxandr [17]3 years ago
5 0

Answer:

<h2>x = 2</h2>

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

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