How do you solve $log_2(x+2) + log_2(x+6)=5$ ?

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Answer 1 · 2016-04-17T01:56:00

$x = 2$

Use $log a + log b = log ab$ and, if $a^x=b, log_a(a^x)=x$ ,

$log_2 (x+2)+log_2(x+6)=log_2((x+2)(x+6))=5$ , with $x+2 >0 and x+5 >0$ . It is sufficient that $x> -2$ .

So, $(x+2)(x+6)=2^5=32$
$x^2+8x-20=0$ .
The roots are $x = 2$ and $x = -10$ .
As $x> -2$ , the root $-10$ is inadmissible. So, $x = 2$ .

How do you solve log_2(x+2) + log_2(x+6)=5log_2(x+2) + log_2(x+6)=5?