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

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Answer 1 · 2018-07-12T15:28:25

#color(blue)(x=2)#

By the laws of logarithms:

#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.