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

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

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Answer 1 · 2018-04-10T04:29:45

approximately:
$x = 2.5468$ $x=2.5468$

Explanation:

${ln}^{\frac{x + 1}{x - 2}}$ $ln^[(x+1)/(x-2)]$ = ${ln}^{x^{2}}$ $ln^(x^2)$
we can cancel out the (Ln) parts and the exponents would be left out;
$\frac{x + 1}{x - 2}$ $(x+1)/(x-2)$ = $x^{2}$ $x^2$
$x + 1$ $x+1$ = $x^{2} . (x - 2)$ $x^2.(x-2)$
$x + 1$ $x+1$ = $x^{3} - 2 x^{2}$ $x^3-2x^2$
$x^{3}$ $x^3$ - $2 x^{2}$ $2x^2$ - $x$ $x$ - $1$ $1$ = $0$ $0$
$x = 2.5468$ $x=2.5468$

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

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Explanation:

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