# pre cal

Approximate the following logarithms using the properties of logarithms given
logb(2)=0.264,
logb(3)=0.419,
and
logb(5)=0.614.

logb(24) =?
logb(100) =?
logb(5b^3) =?

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1. assuming logs base b, we have

log(24) = log(2^3 * 3) = 3log2+log3 = 3*0.264 + 0.419 = 1.211

log(100) = log(10^2) = 2log10 = 2log(2*5) = 2(log2+log5) = ...

log(5b^3) = log5 + 3logb = 0.614 + 3 = 3.614
since logb(b) = 1

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posted by Steve

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