How do you solve #5^(2x)=20#?

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Answer 1 · 2018-05-04T00:56:29

The solution is #x=log_5(20)/2#.

To solve the equation, take the #log_5# of both sides:

#5^(2x)=20#

#log_5(5^(2x))=log_5(20)#

#color(red)cancelcolor(black)(log_5)(color(red)cancelcolor(black)5^(2x))=log_5(20)#

#2x=log_5(20)#

#x=log_5(20)/2#

#color(white)x~~0.930677...#

Hope this helped!