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:

$x=1$