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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!

4 Answers 4

14

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,

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

  • 0
    Thanks for the explanation!2011-04-06
  • 9
    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.)2011-06-20
11

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

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

  • 0
    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 there2011-04-06
  • 0
    Use the change of base identity on both sides $$ \log_b x = \frac{\ln x}{\ln b} = \left( \frac{\ln b}{\ln x} \right)^{-1} = \frac{1}{\log_x b} $$2011-04-07
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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 $$b = \exp \left( \frac{\ln a}{c} \right).$$

  • 0
    @Fabian: Is it conventional to use "exp(...)" rather than "e^(...)"? I noticed that both you and picakhu (in the other answer) did this.2011-04-06
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    @TheChaz: yes. See [this](http://en.wikipedia.org/wiki/Exponential_function#Overview).2011-04-06
  • 2
    @The Chaz: excuse the pun, but it is a TeXnicality.2011-04-06
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    @The Chaz: what is wrong with exp? that is how it reads on the button of my calculator...2011-04-06
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    @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.2011-04-06
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    Thanks, guys. I have noticed some exponential (and fraction-al) expressions being tiny in markup. Wait... was I just referred to the wikipedia page for e^x??? :)2011-04-06
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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