# How do you evaluate the expression 6^(log_6(3x+2))?

Oct 29, 2016

${6}^{{\log}_{6} \left(3 x + 2\right)} = 3 x + 2$

#### Explanation:

Let ${6}^{{\log}_{6} \left(3 x + 2\right)} = u$

then using definition of logarithm that ${\log}_{a} m = b \Rightarrow m = {a}^{b}$, we have

${\log}_{6} u = {\log}_{6} \left(3 x + 2\right)$

i.e. $u = 3 x + 2$

Hence ${6}^{{\log}_{6} \left(3 x + 2\right)} = 3 x + 2$