# Changing base of a logarithm by taking a square root from base?

From my homework I found $$\log_9{x} = \log_3{\sqrt{x}}$$ and besides that an explanation that to this was done by taking a square root of the base. I fail to grasp this completely. Should I need to turn $\log_9{x}$ into base 3, I'd do something like $$\log_9{x} = \frac{\log_3{x}}{\log_3{9}} = \frac{\log_3{x}}{\log_3{3^2}} = \frac{\log_3{x}}{2}$$ but this is a far cry from what I've given as being the correct answer.

Substituting some values to x and playing with my calculator I can see that the answer given as correct is correct whereas my attempt fails to yield the correct answer.

Now the question is, what are correct steps to derive $\log_3{\sqrt{x}}$ from $\log_9{x}$? How and why am I allowed to take a square root of the base and the exponent?

• Write log(9)(x) as log(9)($\sqrt{x}^2)$ and then do the steps which you showed us. – rah4927 Feb 19 '14 at 7:15
• Or remember that $\ \log_a (x^p) \ = \ p \ \log_a (x) \ .$ What is $\ \frac{1}{2} \log_3 x \$ equal to? – colormegone Feb 19 '14 at 7:18

\begin{align}\log_9x=a&\iff\boxed{9^a=x}\\\log_3\sqrt x=b&\iff3^b=\sqrt x\iff\Big(3^b\Big)^2=\Big(\sqrt x\Big)^2\iff3^{2b}=x\iff\boxed{9^b=x}\end{align}\ \bigg\}\to a=b
Note that $a\log b=\log b^a$, so $$\frac{\log_3x}{2}=\frac{1}{2}\log_3x=\log_3x^{1/2}=\log_3\sqrt{x}.$$