How do you solve $2logx = log(-2x+15)$ ?

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Answer 1 · 2015-08-09T23:38:03

$color(red)(x=3)$

$2logx = log(-2x+15)$

Recall that $alog x= log(x^a)$ , so

$logx^2 = log(-2x+15)$

Convert the logarithmic equation to an exponential equation.

$10^(logx^2) = 10^(log(-2x+15))$

Remember that $10^logx =x$ , so

$x^2=-2x+15$

$x^2+2x-15 =0$

$(x+5)(x-3)=0$

$x+5=0$ and $x-3=0$

$x=-5$ and $x=3$

Check:

$2logx = log(-2x+15)$

If $x=-5$ ,

$2log(-5)= log(-2(-5)+15)$

This is impossible, because $log(-5)$ is undefined.

If $x= 3$ ,

$2log3= log(-2×3+15)$

$log3^2= log(-6+15)$

$log9 = log9$

$x=3$ is a solution.

How do you solve 2logx = log(-2x+15)2logx = log(-2x+15)?