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,

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

  • $\begingroup$ Thanks for the explanation! $\endgroup$ – Michael Apr 6 '11 at 23:05
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    $\begingroup$ 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.) $\endgroup$ – 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

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

  • $\begingroup$ 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 $\endgroup$ – Michael Apr 6 '11 at 23:08
  • $\begingroup$ Use the change of base identity on both sides $$ \log_b x = \frac{\ln x}{\ln b} = \left( \frac{\ln b}{\ln x} \right)^{-1} = \frac{1}{\log_x b} $$ $\endgroup$ – 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 $$b = \exp \left( \frac{\ln a}{c} \right).$$

  • $\begingroup$ @Fabian: Is it conventional to use "exp(...)" rather than "e^(...)"? I noticed that both you and picakhu (in the other answer) did this. $\endgroup$ – The Chaz 2.0 Apr 6 '11 at 15:26
  • $\begingroup$ @TheChaz: yes. See this. $\endgroup$ – Willie Wong Apr 6 '11 at 15:29
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    $\begingroup$ @The Chaz: excuse the pun, but it is a TeXnicality. $\endgroup$ – picakhu Apr 6 '11 at 15:30
  • $\begingroup$ @The Chaz: what is wrong with exp? that is how it reads on the button of my calculator... $\endgroup$ – Fabian Apr 6 '11 at 15:33
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    $\begingroup$ @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. $\endgroup$ – Ross Millikan Apr 6 '11 at 15:46

In Excel: to quickly calculate the elusive "e":

To calculate "e" (the base of LN): e = x^(1/LN(x)) Wherein: x = any number >or< 1 but > 0


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