SOLUTION: solve for x
log base 2x + log base 2(x-6)=4
my answer was x=2
am I correct?
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Question 52368: solve for x
log base 2x + log base 2(x-6)=4
my answer was x=2
am I correct?
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
I am having a little difficulty interpretting this, but if I understand what you have typed in, your answer is not right.
log base 2 (x)+log base 2 (x-6)=4
log base 2 ((x)(x-6))=4
(x)(x-6)=2^4
x^2-6x=16
x^2-6x-16=16-16
x^2-6x-16=0
(x-8)(x+2)=0
x-8=0
x-8+8=0+8
x=8
x+2=0
x+2-2=0-2
x=-2
With logs, you have to check for extraneous (false) solutions. Substitute your answers back into the original equation and see if they are true.
For x=8
log base 2 (8)+ log base 2(8-6)=4
log base 2(2^3)+log base 2(2)=4
3+1=4
so x=8 is a solution
for -2, log base 2 (-2) shows that x=-2 is extraneous because you can't take the log of a negative number.
Therefore, the only answer is x=8.
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