Question 451545: solve logbase2(x+3)+ logbase2(x-3)=4
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! solve logbase2(x+3)+ logbase2(x-3)=4
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log2(x+3)+log2(x-3)=4
log2(x=3)(x-3)=4
convert to exponential form: base(2) raised to log of the number(4)=number((x+3)(x-3))
2^4=(x+3)(x-3)
x^2-9=16
x^2=25
x=5
or
x=-5 (reject, (x+3) and (x-3) ≥0
Check:
log2(x+3)+log2(x-3)
=log2(8)+log2(2)
=3+1=4
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