Does log(AB)=LogAxLogB?
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pka
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NO! \(\displaystyle \log(A\cdot B)=\log(A){\Large \bf{+}}\log(B)\)
Dr.Peterson
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If you were working on a test and needed to remember whether this is true, you could check with an example.
Using base ten logs, for example, you should know that [MATH]\log(100) = 2[/MATH] and [MATH]\log(1000) = 3[/MATH]. (Why is that?)
Then, [MATH]\log(100)\log(1000) = (2)(3) = 6[/MATH]; but [MATH]\log(100\cdot1000) = \log(100000) = 5[/MATH]. These aren't equal; so your claim is false.
But you might see from this example that multiplying two powers of ten adds the exponents (that is, here, adds the numbers of zeros: [MATH]2 + 3 = 5[/MATH]); and that might remind you of the correct rule. Since the log is the exponent (that is, [MATH]\log(10^n) = n[/MATH]), this means that the logs of the factors add: [MATH]\log(AB) = \log(A) + \log(B)[/MATH].