Question

como resolver log 3√3 de base 1/9 ?

Answer 1

log 3 $\sqrt{3}$ (1/9)=x

Esse, ^ significa elevado a....

$\frac{1}{9}^x$= 3 $\sqrt{3}$

$\frac{1}{ 3^{2} }$=3 $\sqrt{3}$

3^-2x=3 . 3^1/2
3 ^ -2x=3^(1+1/2)
-2x=3/2
x=-3/4

Answer 2

$log_{\frac{1}{9}} 3\sqrt{3} \\ (\frac{1}{9})^x = 3\sqrt{3}\\ 3^{-2x}= \sqrt{27}\\ 3^{-2x}= \sqrt{3^3}\\ 3^{-2x}= 3^{\frac{3}{2}}\\ -2x= \frac{3}{2} \\ x=\frac{3}{2}: (-2) \\ \\ x= \frac{3}{2} . (- \frac{1}{2}) \\ \\ x= - \frac{3}{4}$