# Log question

1. May 26, 2010

### Everstar

1. The problem statement, all variables and given/known data
1. Use the fact that loga(10)= 2.4197 and loga(13)=2.6955 to solve the following

a. loga(130) b. loga(1.3) c. loga(1000)

2. Relevant equations

3. The attempt at a solution
I'm not really sure how to do this i missed yesterdays class i dont really want the anwser just some help on how to do it becuase i dont understand it out of the book at all. any help would be greatly appreciated

2. May 26, 2010

### Dick

Do you know some rules of logarithms? Like 130=13*10. How is the log of a product related to the log of the terms?

3. May 26, 2010

### Everstar

well i think you getting at something like this
so if i have logam+logan

i get loga(m)(n)

4. May 26, 2010

### Dick

Sure. log_a(m)+log_a(n)=log_a(m*n). So what's the answer to the first one?

5. May 26, 2010

### Everstar

well i'm not sure if im correct but i multiplied the two numbers 2.4197*2.6955 = 6.522

6. May 26, 2010

### Dick

Why did you multiply them? If log_a(m)+log_a(n)=log_a(mn) and you put m=10 and n=13, I don't think you want to multiply 2.4197*2.6955.

7. May 26, 2010

### Everstar

well then i just get log_a(10)(13) which is 130 but i dont understand what i am supposed to do with that? im sorry im just confused

8. May 26, 2010

### Dick

What do you think log_a(10)(13) means? You are a little too confused. Look, log_a(10)+log_a(13)=log_a(10*13)=log_a(130). You know log_a(10)=2.4197 and log_a(13)=2.6955. I wouldn't multiply them.

9. May 26, 2010

### Everstar

oh my goodness lol i add them =)

10. May 26, 2010

### Dick

Thank you. Yes. Now express 1.3 and 1000 in terms of 10 and 13 and use the rules of logs again.

11. May 26, 2010

### Everstar

i figured it out thank you for the help =) i add the first one then divide (subtract the second one, then add add add on the third one. thanks =)

12. May 27, 2010

### HallsofIvy

Yes, log(1000)= log(10*10*10)= log(10)+ log(10)+ log(10). Of course, that's the same as 3*log(10). That's another "rule of logarithms": log(xy)= y log(x).