How do you solve ${(ln x)}^{2} - 1 = 0$ $(lnx)^2 -1 = 0$ ?

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Answer 1 · 2015-12-22T17:41:08

Rearrange then take exponents to find:

$x = e$ $x = e$ or $x = \frac{1}{e}$ $x = 1/e$

Add $1$ $1$ to both sides of the equation to get:

${(ln x)}^{2} = 1$ $(ln x)^2 = 1$

Hence:

$ln x = \pm 1$ $ln x = +-1$

By definition of $ln$ $ln$ we know $e^{ln x} = x$ $e^(ln x) = x$ . Hence:

$x = e^{ln x} = e^{\pm 1}$ $x = e^(ln x) = e^(+-1)$

That is $x = e$ $x = e$ or $x = \frac{1}{e}$ $x = 1/e$

How do you solve (lnx)2−1=0(lnx)^2 -1 = 0?