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

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

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Answer 1 · 2017-10-30T19:39:26

$\frac{3 + \sqrt{13}}{2}$ $(3+sqrt(13))/2$

Explanation:

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

If:

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

Then:

$\frac{x + 1}{x - 2} = x = x^{2} - 3 x - 1$ $(x+1)/(x-2)=x =x^2-3x-1$

Solving: $x^{2} - 3 x - 1$ $x^2-3x-1$

$x = \frac{3 + \sqrt{13}}{2} and x = \frac{3 - \sqrt{13}}{2}$ $x = (3+sqrt(13))/2 and x= (3-sqrt(13))/2$

$\frac{3 - \sqrt{13}}{2}$ $(3-sqrt(13))/2$ not valid as gives negative value in $ln (x - 2)$ $ln(x-2)$

So solution is:

$\frac{3 + \sqrt{13}}{2}$ $(3+sqrt(13))/2$

Answer 2 · 2017-10-30T19:53:28

$x = \frac{3}{2} + \frac{\sqrt{13}}{2}$ $x = 3/2 +sqrt(13)/2$

Explanation:

Taking the exponent of both sides, we get:

$\frac{x + 1}{x - 2} = x$ $(x+1)/(x-2) = x$

Multiplying both sides by $(x - 2)$ $(x-2)$ we get:

$x + 1 = x^{2} - 2 x$ $x+1 = x^2-2x$

Subtract $x + 1$ $x+1$ from both sides to get:

$0 = x^{2} - 3 x - 1$ $0 = x^2-3x-1$

$0 = {(x - \frac{3}{2})}^{2} - \frac{9}{4} - 1$ $color(white)(0) = (x-3/2)^2-9/4-1$

$0 = {(x - \frac{3}{2})}^{2} - {(\frac{\sqrt{13}}{2})}^{2}$ $color(white)(0) = (x-3/2)^2-(sqrt(13)/2)^2$

$0 = (x - \frac{3}{2} - \frac{\sqrt{13}}{2}) (x - \frac{3}{2} + \frac{\sqrt{13}}{2})$ $color(white)(0) = (x-3/2-sqrt(13)/2)(x-3/2+sqrt(13)/2)$

So:

$x = \frac{3}{2} \pm \frac{\sqrt{13}}{2}$ $x = 3/2 +-sqrt(13)/2$

One of these values $\frac{3}{2} + \frac{\sqrt{13}}{2}$ $3/2+sqrt(13)/2$ is positive and a valid solution of the original equation.

The other value $\frac{3}{2} - \frac{\sqrt{13}}{2}$ $3/2-sqrt(13)/2$ is negative so not a solution if $ln$ $ln$ is the real logarithm. In fact it is not even a solution if we consider the complex logarithm, for which:

$ln (\frac{3}{2} - \frac{\sqrt{13}}{2} + 1) - ln (\frac{3}{2} - \frac{\sqrt{13}}{2} - 2)$ $ln(3/2-sqrt(13)/2+1) - ln(3/2-sqrt(13)/2-2)$

$= ln (\frac{5}{2} - \frac{\sqrt{13}}{2}) - ln (\frac{1}{2} + \frac{\sqrt{13}}{2}) - i π$ $= ln(5/2-sqrt(13)/2) - ln(1/2+sqrt(13)/2)-ipi$

$\neq ln (\sqrt{\frac{13}{2}} - \frac{3}{2}) + i π = ln (\frac{3}{2} - \frac{\sqrt{13}}{2})$ $!= ln(sqrt(13/2)-3/2)+ipi = ln(3/2-sqrt(13)/2)$

Answer 3 · 2017-10-30T21:32:29

See below.

Explanation:

In the quest for real solutions, supposing $x \neq {0, 2}$ $x ne {0,2}$

${log}_{e} | x + 1 | - {log}_{e} | x - 2 | = {log}_{e} | x |$ $log_e abs(x+1)-log_e abs(x-2) = log_e absx$ or

${log}_{e} (\frac{| x + 1 |}{| x - 2 | | x |}) = 0$ $log_e(abs(x+1)/(abs(x-2)absx))=0$ or

$\frac{| x + 1 |}{| x - 2 | | x |} = 1$ $abs(x+1)/(abs(x-2)absx)=1$ whose solutions are included in the solutions for

${(x + 1)}^{2} = {(x - 2)}^{2} x^{2}$ $(x+1)^2=(x-2)^2x^2$ or

$(x + 1 - (x - 2) x) (x + 1 + (x - 2) x) = 0$ $(x+1-(x-2)x)(x+1+(x-2)x)=0$ or

$(1 + 3 x - x^{2}) (1 - x + x^{2}) = 0$ $(1+3x-x^2)(1-x+x^2)=0$

and the real solutions are from

$1 + 3 x - x^{2} = 0$ $1+3x-x^2=0$ with the only feasible solution

$x = \frac{1}{2} (3 + \sqrt{13})$ $x = 1/2 (3 + sqrt[13])$

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