How do you solve ${log}_{2} x + {log}_{2} (x + 2) = {log}_{2} (x + 6)$ $log_2x+log_2(x+2)=log_2(x+6)$ ?

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Answer 1 · 2018-07-12T15:28:25

$x = 2$ $color(blue)(x=2)$

By the laws of logarithms:

$log (a) + log (b) = log (a b)$ $log(a)+log(b)=log(ab)$

$:.$

$log_2(x)+log_2(x+2)=log_2(x^2+2x)$

$log_2(x^2+2x)=log_2(x+6)$

$log_a(b)=log_a(c)=>b=c$

$:.$

$x^2+2x=x+6$

$x^2+x-6=0$

Factor:

$(x-2)(x+3)=0=>x=2 and x=-3$

Checking with original equation, we find:

$log_2(-3)$

For real numbers $log(a)$ , $a>0$

So this is undefined.

Therefore only $x=2$ is a solution.

How do you solve log_2x+log_2(x+2)=log_2(x+6)log2x+log2(x+2)=log2(x+6)log_2x+log_2(x+2)=log_2(x+6)?