How do you solve $(2 ln x) + 1 = ln (2 x)$ $(2lnx) + 1 = ln(2x)$ ?

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Answer 1 · 2015-11-30T19:22:45

$x = \frac{2}{e}$ $x = 2/e$

We will use the following:

$2 ln (x) + 1 = ln (2 x)$ $2ln(x) + 1 = ln(2x)$

$\Rightarrow ln (x^{2}) + 1 = ln (2 x)$ $=> ln(x^2) + 1 = ln(2x)$ (by the first property above)

$\Rightarrow ln (x^{2}) - ln (2 x) = - 1$ $=> ln(x^2) - ln(2x) = -1$

$\Rightarrow ln (\frac{x^{2}}{2 x}) = - 1$ $=> ln(x^2/(2x)) = -1$ (by the second property above)

$\Rightarrow ln (\frac{x}{2}) = - 1$ $=> ln(x/2) = -1$

$\Rightarrow e^{ln (\frac{x}{2})} = e^{- 1}$ $=> e^ln(x/2) = e^(-1)$

$\Rightarrow \frac{x}{2} = \frac{1}{e}$ $=> x/2 = 1/e$ (by the third property above)

$:. x = 2/e$

How do you solve (2lnx) + 1 = ln(2x)(2lnx)+1=ln(2x)(2lnx) + 1 = ln(2x)?