How do you solve #lnx+ln(2x)=2#?

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Answer 1 · 2016-08-06T19:04:17

#x = sqrt(2)/2e#

If #x > 0# then:

#ln x + ln(2x) = ln (x*2x) = ln(2x^2)#

and

#2 = ln e^2#

So we have:

#ln (2x^2) = ln (e^2)#

Since #ln# is one-one as a Real valued function of positive Reals, this implies that:

#2x^2 = e^2#

So:

#x^2 = e^2/2#

Hence:

#x = e/sqrt(2) = (sqrt(2))/2e#

Note we ignore the negative square root since we are only looking at the case #x > 0#.