# Does log(AB)=LogAxLogB?

#### Leah5467

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

#### pka

##### Elite Member
My question is Log(AB)=(LogA)(LogB)?
NO! $$\displaystyle \log(A\cdot B)=\log(A){\Large \bf{+}}\log(B)$$

• Leah5467 and topsquark

#### Dr.Peterson

##### Elite Member
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 $$\displaystyle \log(100) = 2$$ and $$\displaystyle \log(1000) = 3$$. (Why is that?)

Then, $$\displaystyle \log(100)\log(1000) = (2)(3) = 6$$; but $$\displaystyle \log(100\cdot1000) = \log(100000) = 5$$. 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: $$\displaystyle 2 + 3 = 5$$); and that might remind you of the correct rule. Since the log is the exponent (that is, $$\displaystyle \log(10^n) = n$$), this means that the logs of the factors add: $$\displaystyle \log(AB) = \log(A) + \log(B)$$.

• Leah5467 and topsquark