How do you solve for x in $log (x+6)=1-log(x-5)$ ?

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Answer 1 · 2016-01-09T13:18:28

$x_1=-(1+sqrt(161))/2$
$x_2=(-1+sqrt(161))/2$

$log(x+6)=1-log(x-5)$
then
$log(x+6)+log(x-5)=1$

using the rule of logaritm product

$log((x+6)(x-5))=1$

Remembering that Field of Existence of $log$ is

$(x+6)(x-5)>0$

we find:

FE: $]-oo,-6[ uu ]5,+oo[$

Now:

$10^(log((x+6)(x-5)))=10^1$

$(x+6)(x-5)=10$

$x^2-5x+6x-30=10$

$x^2+x-40=0$

$x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)$

$x_(1,2)=(-1+-sqrt(1+160))/(2)=(-1+-sqrt(161))/(2)$

$x_1=-(1+sqrt(161))/2$
$x_2=(-1+sqrt(161))/2$

$x_(1,2) in ]-oo,-6[ uu ]5,+oo[$

How do you solve for x in log (x+6)=1-log(x-5)log (x+6)=1-log(x-5)?