How do you solve #e^x=5#?

2 Answers
Gió
Jul 10, 2016

I found: #x=ln5#

Explanation:

You can apply the natural (base #e#) logarithm on both sides:
#ln(e^x)=ln5#
On the left the two operations will cancel out (one the inverse of the other) so you'll get:
#x=ln5#
If you can access a calculator (or tables) you can evaluate it as:
#x=ln5=1.6094#

Jacobi J.
Jul 19, 2018

#x=ln5# or #~~1.609#

Explanation:

We want to cancel out base #e#. We can do this by taking the natural log of both sides.

#lne^x=ln5#

Since #ln# and #e# are inverses of each other, they cancel, and we're left with

#x=ln5#

As a decimal, this is approximately #1.609#.

Hope this helps!

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