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

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Answer 1 · 2016-01-09T13:18:28

$x_{1} = - \frac{1 + \sqrt{161}}{2}$ $x_1=-(1+sqrt(161))/2$
$x_{2} = \frac{- 1 + \sqrt{161}}{2}$ $x_2=(-1+sqrt(161))/2$

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

using the rule of logaritm product

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

Remembering that Field of Existence of $log$ $log$ is

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

we find:

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

Now:

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

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

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

$x^{2} + x - 40 = 0$ $x^2+x-40=0$

$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)$

$x_{1, 2} = \frac{- 1 \pm \sqrt{1 + 160}}{2} = \frac{- 1 \pm \sqrt{161}}{2}$ $x_(1,2)=(-1+-sqrt(1+160))/(2)=(-1+-sqrt(161))/(2)$

$x_{1} = - \frac{1 + \sqrt{161}}{2}$ $x_1=-(1+sqrt(161))/2$
$x_{2} = \frac{- 1 + \sqrt{161}}{2}$ $x_2=(-1+sqrt(161))/2$

$x_{1, 2} \in] - \infty, - 6 [\cup] 5, + \infty [$ $x_(1,2) in ]-oo,-6[ uu ]5,+oo[$

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