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

Question

2889 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-29T10:07:28

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

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

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