Does log(AB)=LogAxLogB?

[Leah5467]

My question is Log(AB)=(LogA)(LogB)? Please help! Thank you!

 

[pka]

My question is Log(AB)=(LogA)(LogB)?

NO! $\displaystyle \log(A\cdot B)=\log(A){\Large \bf{+}}\log(B)$

 

[Dr.Peterson]

My question is Log(AB)=(LogA)(LogB)? Please help! Thank you!

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].