Ratio of logarithms in various bases to other bases

[Kael42]

I'm unsure as to if I am using the correct terminology, but what I mean by this is
log = logarithm in base 10.
ln = logarithm in base e.
logx = logarithm in base x.

Upon some investigation, I found that log(a)/ln(a)=log(b)/ln(b) where a and b are constants,
meaning that there is a ratio between the logarithms.

What is the function of this ratio, in terms of logx and logy?
I.e. If logy(c)=f(x)logx(c), what is f(x)?

The reason behind my search here is to find out how to put a logx (logarithm in base x) function in terms of ln. I want to experiment with various functions in logx on my graphics calculator, but it doesn't have the option to use logarithms in bases other than e and 10.

 

[pwsnafu]

$$\log_{b}(a) = \frac{\log_{d}(a)}{\log_{d}(b)}$$

If you are working with logs have a look at the http://en.wikipedia.org/wiki/List_of_logarithmic_identities" . Very useful

 

[Kael42]

I don't think so. As an example, log(5) does not equal ln(5)/ln(6).

 

[pwsnafu]

I don't think so. As an example, log(5) does not equal ln(5)/ln(6).

$$\log_{10} 5 = 0.6989...$$
$$\frac{ln(5)}{ln(10)} = 0.6989...$$

I have no idea where you got ln(6) from...

 

[Kael42]

My mistake. I missed that b was on both sides. So on the calculator, if I wanted a graph of the logx of 5, I would simply need y=ln5/lnx?
I.e., ln(5-x)?

 

[BobG]

ln5/lnx is not equal to ln(5-x)!

I'm not sure what you're trying to find.

If you just want to be able to find the logarithm of different bases, the change of base formula is what you want.

If you want the ratio between different bases, then the ratio is equal to 1/log b where b is your base. Take the base 10 log of your number and multiply by the ratio. (For example, the natural log of a number is always 2.30 times the base 10 log (plus change - slide rules only go to 3 significant digits and slide rules are one of the main reasons for knowing that ratio).

On a calculator, I'm not sure knowing the ratio will save you any steps.

 

[Kael42]

Thanks BobG, I'm aware of the use of the ratio, I was just wondering what the formula for the ratio was.

As for my error... I blame fatigue. It was late and my head was addled. I messed up my log laws.