Question

expand log 243 to the base 3 root 3

Answer 1

$\textbf{Concept used:}$

$\textbf{Logarithmic notation}$

$\bf\;a^x=N\implies\;log_aN=x$

$\text{Let, }log_{3\sqrt3}243=x$

$\implies(3\sqrt3)^x=243$

$\implies(3\sqrt3)^x=3^5$

$\implies((\sqrt3)^3)^x=((\sqrt3)^2)^5$

$\implies(\sqrt3)^{3x}=(\sqrt3)^{10}$

$\text{Equating powers on both sides, we get}$

$3x=10$

$\implies\boxed{\bf\;x=\frac{10}{3}}$

Find more:

log2 log2 log2 X=1

https://brainly.in/question/6856970#

Answer 2

We need to know that

$log_m(m^n) = n$

$log_m(ab) = log_ma+log_mb$

This is the concept applied in the given problem

We have

$243 = 3^5 = (3\sqrt3)^3*\sqrt3$

Hence,  

$log_{3\sqrt3}(243) = log_{3\sqrt3}(3\sqrt3)^3+log_{3\sqrt3}\sqrt3$

$= 3+\frac{1}{3}$

$= \frac{10}{3}$

HOPE IT HELPS!!