Question

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

Answer 1

`6^(log_6(3x+2))=3x+2`

Explanation:

Let `6^(log_6(3x+2))=u`

then using definition of logarithm that `log_a m=b rArr m=a^b`, we have

`log_6 u=log_6(3x+2)`

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

Hence `6^(log_6(3x+2))=3x+2`