logarithmic equation

Algebra Level 1

If $\log_{8}m + \log_{8}\frac{1}{6}=\frac{2}{3}$ , then what is $m?$

2 solutions

Akshat Sharda
Sep 29, 2015

$\Rightarrow \log_{8}m + log_{8}\frac{1} {6}=\frac{2}{3}$

$\Rightarrow \log_{8}\frac{m}{6}=\frac{2}{3}$

$\Rightarrow \frac{m}{6}=8^{\frac{2}{3}}$

$\Rightarrow m=\boxed{24}$

i didnt understand the 3 rd step

abhishek alva - 4 years, 12 months ago

its an identity

Shreyansh Tiwari - 4 years, 11 months ago

what is that identity

abhishek alva - 4 years, 11 months ago

@Abhishek Alva – $a^x=N\Leftrightarrow \log_{a}N=x$

Akshat Sharda - 4 years, 11 months ago

@Akshat Sharda – i knew this identity but couldnt use at right place

abhishek alva - 4 years, 11 months ago

This is wrong

Joshua Productions - 3 years, 8 months ago

I didn't understand the 3rd step

Gourango Tanti - 3 years, 3 months ago

it's wrong

Ionut Ghinghes - 3 years, 3 months ago

This is wrong.

Kirankumar Sanap - 2 years, 11 months ago

This is wrong. You just substitute the value and solve

Nimith Kanangot - 2 years, 10 months ago

Sorry for wrong posting this is correct.logx to base a +log y to base a is not equal to log x+y to base a .I accidently made that mistake

Nimith Kanangot - 2 years, 10 months ago

A Former Brilliant Member
Mar 24, 2019

$\log_8m+\log_8\frac{1}{6}=\frac{2}{3}$

$\log_8[m\times \frac{1}{6}]=\frac{2}{3}$

$8^\frac{2}{3}=\frac{m}{6}$

$m=6\times8^\frac{2}{3}=\large{\boxed{24}}$