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-Dominant- [34]
3 years ago
6

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

Mathematics
2 answers:
Charra [1.4K]3 years ago
5 0
\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
Ann [662]3 years ago
4 0
X=2  hope this helps!!!! ^_^
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Answer:

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The solution is (-3, -2).

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If you'd like to solve the set of equations, substitution for x works nicely.

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3 years ago
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