How do you solve #log5^(2x) = 4#?

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Answer 1 · 2015-11-04T17:39:17

The solution is #x=2/log(5)#

A property of lohgarithms states that

#log(a^b)=blog(a)#

In your case, #a=5# and #b=2x#, so the equality becomes

#log(5^{2x})=2xlog(5)#.

At this point, it's very easy to isolate the #x#, since #log(5)# is just a number, and we have

#log(5^{2x}) = 4#

#iff#

#2xlog(5)=4#

#iff#

#x=4/(2log(5)) = 2/log(5)#