How do you solve $log_2 x+5=8-log_2 (x+7)$ ?

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Answer 1 · 2016-04-27T12:40:00

$x_1=-9,61$
$x_2=2,61$

$log_2x+5=8-log_2(x+7)$

$log_2x+log_2(x+7)=8-3$

$log_ab+log_a c=log_a(b*c)$

$log_2 x+log_2(x+7)=log_2 x(x+7)$

$log_2 x(x+7)=5$

$log_a b=k$

$b=a^k$

$x(x+7)=2^5$

$x^2+7x-25=0$

$Delta=sqrt(7^2+4*1*25)$

$Delta=sqrt(49+100)" "Delta=sqrt(149)" "Delta=12,21$

$x_1=(-7-12,21)/2" "x_1=-9,61$

$x_2=(-7+12,21)/2" "x_2=2,61$

How do you solve log_2 x+5=8-log_2 (x+7)log_2 x+5=8-log_2 (x+7)?