How do you solve for x in $2 (log x - log 6) = log x - 2 log \sqrt{x - 1}$ $2(logx-log6)=logx-2logsqrt( x - 1)$ ?

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How do you solve for x in $2 (log x - log 6) = log x - 2 log \sqrt{x - 1}$ $2(logx-log6)=logx-2logsqrt( x - 1)$ ?

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Answer 1 · 2016-02-03T15:00:34

$x = \frac{1 + \sqrt{145}}{2} \approx 6.52$ $x=(1+sqrt(145))/2 ~~6.52$
(at least that's what I got)

Explanation:

Given
$XXX 2 (log (x) - log (6)) = log (x) - 2 log (\sqrt{x - 1})$ $color(white)("XXX")color(red)(2(log(x)-log(6))) = color(blue)(log(x)-2log(sqrt(x-1)))$

Thinks that are helpful to know:
[1] $XXX k \cdot log (a) = log (a^{k})$ $color(white)("XXX")k*log(a) = log(a^k)$
[2] $XXX log (a) - log (b) = log (\frac{a}{b})$ $color(white)("XXX")log(a)-log(b)=log(a/b)$
[3] $XXX log (\sqrt{a}) = \frac{log (a)}{2}$ $color(white)("XXX")log(sqrt(a))=log(a)/2$ (this actually follows from [1])
[4] $XXX$ $color(white)("XXX")$ for $log (a)$ $log(a)$ to be meaningful $a > 0$ $a>0$ .

$2 (log (x) - log (6))$ $color(red)(2(log(x)-log(6)))$

$XXX = 2 (log (\frac{x}{6}))$ $color(white)("XXX")=color(red)(2(log(x/6))$ from [2]

$XXX = log (\frac{x^{2}}{6^{2}})$ $color(white)("XXX")=color(red)(log((x^2)/(6^2))$ from [1]

$log (x) - 2 log (\sqrt{x - 1})$ $color(blue)(log(x)-2log(sqrt(x-1)))$

$XXX = log (x) - 2 (\frac{log (x - 1)}{2})$ $color(white)("XXX")=color(blue)(log(x)-2(log(x-1)/2))$ from [3]

$XXX = log (x) - log (x - 1)$ $color(white)("XXX")=color(blue)(log(x)-log(x-1))$ simplification

$XXX = log (\frac{x}{x - 1})$ $color(white)("XXX")=color(blue)(log(x/(x-1))$ from [2]

Therefore
$XXX log (\frac{x^{2}}{6^{2}}) = log (\frac{x}{x - 1})$ $color(white)("XXX")color(red)(log(x^2/(6^2))) = color(blue)(log(x/(x-1)))$

$\Rightarrow$ $rArr$ $XXX \frac{x^{2}}{36} = \frac{x}{x - 1}$ $color(white)("XXX")x^2/36 = x/(x-1)$

$\Rightarrow$ $rArr$ $XXX x^{3} - x^{2} = 36 x$ $color(white)("XXX")x^3-x^2=36x$

$\Rightarrow$ $rArr$ $XXX x (x^{2} - x - 36) = 0$ $color(white)("XXX")x(x^2-x-36)=0$

$\Rightarrow$ $rArr$ $XXX x = 0 XX or XX (x^{2} - x - 36) = 0$ $color(white)("XXX")x=0color(white)("XX")orcolor(white)("XX")(x^2-x-36)=0$

But $x \neq 0$ $x!=0$ from [4]
So
$XXX (x^{2} - x - 36) = 0$ $color(white)("XXX")(x^2-x-36)=0$

This can be factored using the quadratic formula to get
$XXX x = \frac{1 + \sqrt{145}}{2} XX or XX x = \frac{1 - \sqrt{145}}{2}$ $color(white)("XXX")x=(1+sqrt(145))/2color(white)("XX")orcolor(white)("XX")x=(1-sqrt(145))/2$

But $\frac{1 - \sqrt{145}}{2} < 0$ $(1-sqrt(145))/2 < 0$ so $x \neq \frac{1 - \sqrt{145}}{2}$ $x!=(1-sqrt(145))/2$ from [4]

Therefore
$XXX x = (1 + \frac{\sqrt{145}}{2}) \approx 6.52$ $color(white)("XXX")x=(1+sqrt(145)/2)~~6.52$

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