How do you find log 5 when given log 2 and log 3?

Question

logs are in base ten, by the way.

Answer 1

You'll have to use log 10, which is trivial because it's equal to 1. Remember that log a/b = log a - log b, so log 5 = log 10/2 = log 10 - log 2 = 1 - log 2. I didn't need to use log 3.

Answer 2

There are a lot of wrong answers here. The basic log formula is log(ab)=log(a) +log(b). Since log(10)=1 and 2*5=10, we get that
1=log(10)=log(2*5)=log(2)+log(5).
Since you have log(2), you can solve for
log(5)=1-log(2).
The log of 3 is not needed here.
If you were doing natural logs, there would be no way to find log(5) directly. You need to be working base 10, not base e.

Answer 3

log(a)*log(b)=log(a+b)

Therefore:
log2*log3=log(2+3)=log(5)

So to find log5, multiply log2 by log3

(It is also true that log(a)/log(b)=log(a-b), but that doesn't use both log2 and log3, and therefore isn't the method which is being looked for.)

Answer 4

here i am working with natural logs

5=e^(ln5), 3=e^(ln3), 2=e^(ln2)

(2+3)= e^(ln2)+e^(ln3)

ln(2+3)=ln(e^(ln2)+e^(ln3))

ln(5) =ln(e^(ln2)+e^(ln3)) (relationship)

you can't manipulate logs like ordinary numbers

Answer 5

log5=log10/2=log10-log2
=1-.3010=.6990

Answer 6

lg 5 = (lg 10)/(lg 2)

Hope this helps.

Answer 7

I agree with DavidK !