How do you solve $ln (x^{2} + 12) = ln x + ln 8$ $ln(x^2+12)=lnx+ln8$ ?

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How do you solve $ln (x^{2} + 12) = ln x + ln 8$ $ln(x^2+12)=lnx+ln8$ ?

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Answer 1 · 2016-10-17T09:49:36

$x = 2$ $x=2$ and $x = 6$ $x=6$

Explanation:

$ln (x^{2} + 12) = ln x + ln 8$ $ln(x^2+12)=lnx+ln8$
We can rewrite this as $ln (x^{2} + 12) = ln 8 x$ $ln(x^2+12)=ln8x$
Removing the ln on both sides,give
$x^{2} + 12 = 8 x$ $x^2+12=8x$
Rearranging
$x^{2} - 8 x + 12 = 0$ $x^2-8x+12=0$
$(x - 6) (x - 2) = 0$ $(x-6)(x-2)=0$
so x=6 or x=2
We have to check the conditions for ln
$x^{2} + 12 > 0$ $x^2+12>0$ and $8 x > 0$ $8x>0$
So the solution is ok

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How do you solve $ln (x^{2} + 12) = ln x + ln 8$ $ln(x^2+12)=lnx+ln8$ ?

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How do you solve $ln (x^{2} + 12) = ln x + ln 8$ $ln(x^2+12)=lnx+ln8$ ?