Confused over application of log power rule ?

The Question says

In x+In2x=12
I got the right answer by e^(2x^2)=e^12 which gives 218.39
But when I try it with log power rule,
Inx+In2x=12
In2x^2=12
2In2x=12
In2x=6
e^(In2x)=e^6,which gives me the wrong answer of 201.7. Why is this ?

1 Answer
Eddie
Aug 14, 2016

Both your answers are wrong. See below.

Correct is #e^6 / sqrt 2 approx 285.27#

Explanation:

#ln x+ln2x=12 #

I got the right answer by #color(red)(e)^(2x^2)=e^12# which gives 218.39

No, red bit is wrong

#ln x+ln2x=12 #
#implies ln 2x^2=12 #
#implies e^(ln 2x^2)=e^12 #
#implies 2x^2=e^12 # NOT #e^(2x^2)=e^12#
#implies x^2=e^12 /2 #
#implies x=(e^12 /2)^(1/2) = e^6 / sqrt 2 #

But when I try it with log power rule,
#lnx+ln2x=12#
#ln2x^2=12#
#2ln2x=12# Nah, sorry

Correct is
#ln2x^2=12#
#ln(sqrt 2 x)^2=12#
#2 ln(sqrt 2 x)=12#
# ln(sqrt 2 x)=6#
# e^(ln(sqrt 2 x))=e^6#
# sqrt 2 x=e^6#
# x=e^6 / sqrt 2 #

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