How do you solve log_3 x – 1 = 3?

May 8, 2015

Ans: $28$

I think that it is ${\log}_{3} \left(x - 1\right) = 3$
So you can write:
${3}^{{\log}_{3} \left(x - 1\right)} = {3}^{3}$
cancelling the $3$ and the $\log$ on the left you get:
$x - 1 = 27$
$x = 28$

The question could also be ${\log}_{3} \left(x\right) - 1 = 3$ (with only $x$ in the argument although I doubt it) so you get:
${\log}_{3} \left(x\right) = 3 + 1$
${\log}_{3} \left(x\right) = 4$ in the same way as before:
${3}^{{\log}_{3} \left(x\right)} = {3}^{4}$
$x = 81$