Here is an example problem where the logarithm is expressed as an exponent. Please help me understand this concept its not properly covered in my textbook. $$ 3^{\log_3 (2k)} = 9 $$
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$\begingroup$HINT: taking the logarithm on both sides we get $$\log_{3}{2k}\ln(3)=\ln(3^2)$$ and this is equal to $$\log_{3}{2k}=2$$ from here we get $$3^2=2k$$ Can you finish?
$\endgroup$ 3 $\begingroup$By definition, $a^{\log_a x}=x$, as $\log_a x$ is the unique number $b$ such that $a^b=x$.
Thus the equation is simply $2k=9$.
I wouldn't use the longer way of writing the equation as $3^{\log_3(2k)}=3^2$, from which we can infer $\log_3(2k)=2$ and therefore $2k=9$ by definition of logarithm.
$\endgroup$ $\begingroup$Whenever there are logarithms in equations you usually come back to the rules of logarithms. Namely :
$\,\ln(a^b)=b\ln(a) \\a^{log_ax}=x \\ log_a(x)=b \implies a^b=x\\$
Now in your question $\,3^{\log_3 (2k)} = 9 \\2k = 9\\k=\frac92$
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