## Shido88 Solve for x 83=27^x Consider that: log(a^x)=x (times) log(a) one year ago one year ago

1. abb0t

It looks like you already have your answer there. So just apply it and solve for x.

2. Mertsj

$\log(83)=xlog(27)$ $x=\frac{\log(83)}{\log(27)}$

3. Shido88

on the log(27) part, do you just ignore the x over the 27 ? or where did it goes ?

4. Mertsj

That's where the rule comes in. if 8=2^3 then log8=3log2