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 $x > 0$ .

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