Question

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

Answer 1

$\bf \textit{Logarithm of rationals} \\\\ log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y) \\\\\\ \textit{Logarithm of exponentials} \\\\ log_a\left( x^b \right)\implies b\cdot log_a(x)\\\\ -------------------------------$

$\bf log_2(2x^3-8)-2log_2(x)=log_2(x) \\\\\\ log_2(2x^3-8)-log_2(x^2)=log_2(x)\implies log_2\left( \cfrac{2x^3-8}{x^2} \right)=log_2(x) \\\\\\ \textit{since both sides are getting }log_2\textit{ we can simply undo that} \\\\\\ \cfrac{2x^3-8}{x^2}=x\implies 2x^3-8=x^3\implies x^3-8=0 \\\\\\ x^3=8\implies x=\sqrt[3]{8}\implies x=2$

Answer 2

X=2  hope this helps!!!! ^_^