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?

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

Beginner2022-06-21Added 20 answers

You are only one step away!

Note that $a\mathrm{log}b=\mathrm{log}{b}^{a}$, so

$\frac{{\mathrm{log}}_{3}x}{2}=\frac{1}{2}{\mathrm{log}}_{3}x={\mathrm{log}}_{3}{x}^{1/2}={\mathrm{log}}_{3}\sqrt{x}.$

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

Beginner2022-06-22Added 5 answers

$\begin{array}{rl}{\mathrm{log}}_{9}x=a& \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\overline{){\displaystyle {9}^{a}=x}}\\ {\mathrm{log}}_{3}\sqrt{x}=b& \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{3}^{b}=\sqrt{x}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}{3}^{b}{{\textstyle )}}^{2}={\textstyle (}\sqrt{x}{{\textstyle )}}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{3}^{2b}=x\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\overline{){\displaystyle {9}^{b}=x}}\end{array}\text{}{\textstyle \}}\to a=b$