crossoverman9b

2022-06-20

Changing base of a logarithm by taking a square root from base?
From my homework I found
${\mathrm{log}}_{9}x={\mathrm{log}}_{3}\sqrt{x}$
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 ${\mathrm{log}}_{9}x$ into base 3, I'd do something like
${\mathrm{log}}_{9}x=\frac{{\mathrm{log}}_{3}x}{{\mathrm{log}}_{3}9}=\frac{{\mathrm{log}}_{3}x}{{\mathrm{log}}_{3}{3}^{2}}=\frac{{\mathrm{log}}_{3}x}{2}$
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 ${\mathrm{log}}_{3}\sqrt{x}$ from ${\mathrm{log}}_{9}x$? How and why am I allowed to take a square root of the base and the exponent?

mallol3i

You are only one step away!
Note that $a\mathrm{log}b=\mathrm{log}{b}^{a}$, so
$\frac{{\mathrm{log}}_{3}x}{2}=\frac{1}{2}{\mathrm{log}}_{3}x={\mathrm{log}}_{3}{x}^{1/2}={\mathrm{log}}_{3}\sqrt{x}.$

Yesenia Sherman