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

Question

1221 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-06T16:31:17

Applying the exponential on both sides.

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#