# How do I find the base when Log is given

I'm trying to figure out how to calculate the base if:

$$\log_b 30 = 0.30290$$

How do I find $b$ ?

I've slaved over the Wikipedia page for logarithms, but I just don't get the mathematical notations.

If someone could let me know the steps to find $b$ in plain english, I'd be eternally grateful!

-

You need to think about the definitions.

Since $a^b=c$ can be rewritten as $\log_a c = b$.

That should tell you that,

$$b^{0.30290} = 30$$

and then,

$$b = \exp {\frac{\ln 30}{0.30290}}$$

-
Thanks for the explanation! –  Michael Apr 6 '11 at 23:05
After your first equation $b^{0.30290} = 30$, it would IMHO be more straightforward to raise both sides to the power $1/0.30290$ and write $b = 30^{1/0.30290}$. (For those who don't know the $\exp$ and $\ln$ functions but have some middle-school level understanding of exponentiation.) –  ShreevatsaR Jun 20 '11 at 11:10

The change-of-base identity says the following: fixing $\ln$ to mean the natural logarithm (logarithm with base $e$), $$\log_b x = \frac{\ln x}{\ln b}$$ and as a consequence, you can derive the statement that $$\log_b x = \frac{1}{\log_x b}.$$

This tells you that your statement $$\log_b 30 = 0.30290$$ is equivalent to $$\log_{30} b = \frac{1}{0.30290}$$ so that

$$b = 30^{\frac{1}{0.30290}} \sim 75265.70$$

-
Thank you very much for the help. I have my answer now so I should really just fill in the blanks and not think about the 'why', otherwise I'll go mad ... BUT .. why is lnx/lnb == 1/logxb .. you totally lost me there –  Michael Apr 6 '11 at 23:08
Use the change of base identity on both sides $$\log_b x = \frac{\ln x}{\ln b} = \left( \frac{\ln b}{\ln x} \right)^{-1} = \frac{1}{\log_x b}$$ –  Willie Wong Apr 7 '11 at 10:31
Once you have log of one base (e.g. the natural log $\ln$), you can easily calculate the log of any basis via $$\log_b a = \frac{\ln a}{\ln b}.$$
In your case you want to solve $\log_b a =c$ for $b$, which is easily done using the formula above with the solution $$\ln b = \frac{\ln a}{c}$$ or equivalently $$b = \exp \left( \frac{\ln a}{c} \right).$$
@The Chaz: two advantages of $\exp(\ldots)$ are that it keeps the exponent larger and easier to read, and that $e$ on this page is sometimes ugly (it is a funny script e-I don't know what controls it). But people use both. –  Ross Millikan Apr 6 '11 at 15:46