How do you solve #5^ { 2x } + 20( 5^ { x } ) - 125= 0#?

1 Answer
Noah G
Mar 3, 2017

#x = 1#.

Explanation:

The equation can be written as

#(5^x)^2 + 20(5^x) - 125 = 0#

We now let #u = 5^x#.

#u^2 + 20u - 125 = 0#

#(u + 25)(u - 5) = 0#

#u = -25 and 5#

#5^x = -25 and 5^x = 5#

There is no solution to the first solution (because if you take the #ln# of #-25#, it is undefined). The second equation obviously has #x = 1# as a solution.

Hopefully this helps!

Related questions
See all questions in Natural Logs
Impact of this question
3579 views around the world
You can reuse this answer
Creative Commons License