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

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Answer 1 · 2015-08-09T23:38:03

$x = 3$ $color(red)(x=3)$

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

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

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

Convert the logarithmic equation to an exponential equation.

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

Remember that $10^{log x} = x$ $10^logx =x$ , so

$x^{2} = - 2 x + 15$ $x^2=-2x+15$

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

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

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

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

Check:

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

If $x = - 5$ $x=-5$ ,

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

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

If $x = 3$ $x= 3$ ,

$2 log 3 = log (- 2 \times 3 + 15)$ $2log3= log(-2×3+15)$

${log 3}^{2} = log (- 6 + 15)$ $log3^2= log(-6+15)$

$log 9 = log 9$ $log9 = log9$

$x = 3$ $x=3$ is a solution.

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