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

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Answer 1 · 2015-09-18T15:32:11

$x = 20$

Put everything that's a log on the same side
$log(x-15)+log(x) = 2$

Remember that $log(m) + log(n) = log(mn)$
$log(x(x-15)) = 2$

If $log_a(b) = c$ , then $b = a^c$
$x(x-15) = 10^(2)$

Expand and solve the quadratic equation

$x^2 - 15x = 100 rarr x^2 -15x -100 = 0$
$x = (15 +-sqrt(225 - 4*1(-100)))/2 = (15 +-sqrt(225 +400))/2$
$x = (15 +-sqrt(625))/2 = (15 +-25)/2$

$x_1 = (15+25)/2 = 40/2 = 20$
$x_2 = (15-25)/2 = -10/2 = -5$

Remember that since we were dealing with logarithms, we can't have null or negative arguments, so
$x-15 > 0 rarr x > 15$
$x > 0$

We conclude that any answers must follow $x > 15$ , which only of the two answers do, thus, the answer is $x = 20$

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