Question #645d2

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Answer 1 · 2017-07-06T20:05:36

I got: $x = 14.09017$ $x=14.09017$

I would use a property of logs as:
#logx+logy=log(xy)3 and write:

${log}_{5} [(x - 11) (x - 6)] = 2$ $log_5[(x-11)(x-6)]=2$

then use the definition of log to write:

$(x - 11) (x - 6) = 5^{2}$ $(x-11)(x-6)=5^2$

$(x - 11) (x - 6) = 25$ $(x-11)(x-6)=25$

rearrange:

$x^{2} - 6 x - 11 x + 66 - 25 = 0$ $x^2-6x-11x+66-25=0$

$x^{2} - 17 x + 41 = 0$ $x^2-17x+41=0$

Solve using the Quadratic Formula:

$x_{1, 2} = \frac{17 \pm \sqrt{289 - 164}}{2} =$ $x_(1,2)=(17+-sqrt(289-164))/2=$

$x_{1} = 14.09017$ $x_1=14.09017$
$x_{2} = 2.90903$ $x_2=2.90903$

$x_{2}$ $x_2$ is small and would make the arguments of the original logs negative.