Question #158af

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Answer 1 · 2017-10-21T12:01:56

#x=1#

#ln(a)+ln(b)= ln(ab)#

So:

#ln(x^3) + ln(2/x)= ln((2x^3)/x)=ln(2x^2)#

If:

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

Then:

#2x^2=2=>x=+-sqrt(1)#

Only #x=1# is valid, since logarithms of negative numbers are undefined.

As a note:

You could have solved this just using the clue you were given.

#2lnx=0#

#lnx=0#

This means that if e is raised to the power of 0 it equals x.

#e^0=x#

#e^0=1#

So:

#x=1#