Can you distribute logarithms?

1. Aug 11, 2013

mileena

Does anyone know if the following is true:

logb (x + y) = logb x + logb y

Thanks. This isn't homework, but I am just wondering if the following is true. I already know the logarithm product, quotient, and power rules!

Last edited: Aug 11, 2013
2. Aug 11, 2013

micromass

Staff Emeritus
Why don't you try to find a counterexample?

3. Aug 11, 2013

mileena

Ok, good idea. I don't know how to use a scientific calculator yet to figure out logs, but I know they are online. I will plug in some real numbers to see if the equation s true or not.

4. Aug 11, 2013

mileena

Ok, I just did, using log 2 and log 3, and then again with log 8 and log 10. It worked! you can distribute logs!

Thanks, and sorry for posting so many questions in one day.

5. Aug 11, 2013

micromass

Staff Emeritus
It shouldn't work

Can you post what you did?

6. Aug 11, 2013

junaid314159

7. Aug 11, 2013

johnqwertyful

log(2)=log(1+1)=log(1)+log(1)

Is this true?

8. Aug 11, 2013

Staff: Mentor

What's the log of 1?
What's the log of 2?

9. Aug 11, 2013

micromass

Staff Emeritus
I think he knows lol. He just asked it to the OP :tongue:

10. Aug 11, 2013

Staff: Mentor

Oops - right you are. Sorry about that.

11. Aug 11, 2013

symbolipoint

Make and test an example. Imagine your base is 10, and x=100 and y=10,000,000.

log10(100+10000000)=log10(100)+log10(10000000)
Does this make sense? Does this not make sense?

12. Aug 12, 2013

mathman

Log(xy) = log(x) + log(y). Unless xy = x+y, your equation is wrong.

13. Aug 12, 2013

mileena

Hi, sorry for not posting sooner, but I was busy all day and I don't really have Internet access until I get to the library.

Let me also say that I am an idiot!

Yesterday, I said that you could distribute logs, so that:

logb (x + y) = logb (x) + logb (y)

But I made a mistake. Instead of adding, for example, 2 and 3, and taking the log of 5 and comparing that with the sum of log 2 and log 3, I multiplied 2 and 3! Thus I got:

logb x + logb y = logb (xy)

which is, of course, the product rule.

14. Aug 12, 2013

symbolipoint

I did not say that the equation was correct. I only presented it, and then asked two questions. I know already that the equation is wrong. mileena already found understanding that was sought.

15. Aug 12, 2013

junaid314159

@symbolipoint: I think mathman may have been responding to the original post when he was commenting.

@Nugatory: log 1 = 0 and log 2 ≈ 0.6931...

-Junaid :tongue:

16. Aug 13, 2013

symbolipoint

mathman, that is possible. I inferred that you may have responded to my post because yours came directly after it, and other interrelations of responses of posts were not clear.

I tried to give an example to be checked and asked if the example made sense.

17. Aug 13, 2013

mathman

I was responding to the original post.