log(x^n)=n.log(x)

\(
Prove\ log_b(x^n)=nlog_b(x)\\
log_b(x^n)=m\Rightarrow b^m=x^n\Rightarrow x=b^{\frac{m}{n}}\\
n\cdot log_b(x) = p\Rightarrow log_b(x)=\frac{p}{n}\Rightarrow x=b^\frac{p}{n}\\
Equating\ p=m \rightarrow log_b(x^n)=nlog_b(x)
\)

 

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