How to solve equations when logarithm is the exponent?

Question

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. $$ 3^{\log_3 (2k)} = 9 $$

Answer 1

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?

Answer 2

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.

Answer 3

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 $\,3^{\log_3 (2k)} = 9 \\2k = 9\\k=\frac92$