Had a question someone might be able to answer...
Why do I get the same number if I divide, let's say log5/log3, as if I divided ln5/ln3???
What exactly is the relationship between logs and natural logs that makes this true?
Thanks in advance.
M
Had a question someone might be able to answer...
Why do I get the same number if I divide, let's say log5/log3, as if I divided ln5/ln3???
What exactly is the relationship between logs and natural logs that makes this true?
Thanks in advance.
M
Those better not yield the same answer.
log(5)=x where x satisfies $\displaystyle 10^x=5$
ln(5)=x where x satisfies $\displaystyle e^x=5$
log(3)=x where x satisfies $\displaystyle 10^x=3$
ln(3)=x where x satisfies $\displaystyle e^x=3$
If you got the same answer when you calculated these ratios you have made an error somewhere.
If you are interested, you can express logs in different bases in terms of the natural log as follows.
$\displaystyle y=log_a(b)\Rightarrow a^y=b\Rightarrow ln(a^y)=ln(b)\Rightarrow yln(a)=ln(b) \Rightarrow y=\frac{ln(b)}{ln(a)}$
So we see that
$\displaystyle log_a(b)=\frac{ln(b)}{ln(a)}$
hmm now that is strange. generally log without any subscript is understood to be log base 10. Whereas ln is to be log base e. might be worth seeing whether it is giving you ln or log though so you can apply the change of base stuff as necessary. try log(10) and ln(e) if you guess the base right you should get 1. That oughta at least tell you which base it is giving you.
I thought so too, but we did several examples to prove what I don't know, that's what I am trying to figure out...lol.
I have been curious since Monday and I have tried researching all I can about logarithms. I don't have either of my calculators on me but I am going to keep trying to figure it out.
Thanks for your help though. If you have a calculator handy, input the numbers, first using the log key then use the ln key. Weird.
oh dude, yeah you are right they should be the same, I was thinking $\displaystyle ln(5/3)$ and $\displaystyle log(5/3)$ my bad dude. Here is why it is true.
$\displaystyle a=log(5)\Rightarrow 10^a=5 \Rightarrow a=\frac{ln(5)}{ln(10)}$
$\displaystyle b=log(3)\Rightarrow 10^b=5 \Rightarrow b=\frac{ln(3)}{ln(10)}$
$\displaystyle \frac{log(5)}{log(3)}=\frac{a}{b}=\frac{\frac{ln(5 )}{ln(10)}}{\frac{ln(3)}{ln(10)}}=\frac{ln(5)}{ln( 3)}$
It's the change-of-base formula:
log_b(x) = log_c(x)/log_c(b)
This is because:
log_b(x) = y
x = b^y
log_c(x) = log_c(b^y) = y*log_c(b)
log_c(x)/log_c(b) = y = log_b(x)
By applying the change-of-base formula twice (once backwards and once forwards), we get:
log_10(5)/log_10(3) = log_3(5) = ln(5)/ln(3)