How do you solve #Ln x^2 = 2#?

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Answer 1 · 2016-07-02T15:12:40

The solution is #x=e#.

First you apply the rule of the log for the powers

#ln(x^k)=kln(x)# then, for you it is

#ln(x^2)=2#

#2ln(x)=2#

#ln(x)=1#.

Then, to obtain #x# you need to apply the inverse operation of #ln# that is the exponential

#e^ln(x)=e^1#

#x=e#.