How do you solve $e^{ln x - 3 ln x} = (\frac{5}{2})$ $e ^(lnx-3lnx) = (5/2)$ ?

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Answer 1 · 2016-04-17T10:09:14

$x = \sqrt{\frac{2}{5}}$ $x=sqrt(2/5)$

Note that $x$ $x$ cannot be negative.

Taking log of both sides to base $e$ $e$ we get

$ln x - 3 ln x = ln (\frac{5}{2})$ $lnx-3lnx=ln(5/2)$ now simplifying LHS

$ln x - 3 ln x = ln x - {ln x}^{3} = ln (\frac{x}{x^{3}}) = ln (\frac{1}{x^{2}}) = ln (\frac{5}{2})$ $lnx-3lnx=lnx-lnx^3=ln(x/x^3)=ln(1/x^2)=ln(5/2)$

Hence $\frac{1}{x^{2}} = \frac{5}{2}$ $1/x^2=5/2$ or $x^{2} = \frac{2}{5}$ $x^2=2/5$

Hence $x = \pm \sqrt{\frac{2}{5}}$ $x=+-sqrt(2/5)$ but x cannot be negative.

Hence $x = \sqrt{\frac{2}{5}}$ $x=sqrt(2/5)$ .

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