How do you solve #log(x-5)=-2#?

1 Answer
Jun 6, 2016

Answer:

Applying the exponential on both sides.

Explanation:

Your #x# is in the #log#, so you have to remove it. The inverse operation of the #log# is the exponential, so you can do

#log(x-5)=-2#

#e^(log(x-5))=e^-2#

Because, as said, the exponential is the inverse of the logarithm, we have

#x-5=e^-2#

and finally

#x=e^-2+5\approx 5.13#