Question 177517
log base of 2 times (x+20) minus log base of 2 times (x+2) is equal to log base of 2 x
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log(base2)(x+20) - log(base2)(x+2) = log(base2)x
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It's all base 2 so just keep that in mind:
log(x+20) - log(x+2) = log2
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since logA - logB = log(A/B) you get:

log[(x+20)/(x+2)] = log2

Taking the inverse log you get:
(x+20)/(x+2) = 2
x+20 = 2(x+2)
x + 20 = 2x + 4
x = 16
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Cheers,
Stan H.