18

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!

Willie Wong
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Michael
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5 Answers5

18

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,

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

picakhu
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  • Thanks for the explanation! – Michael Apr 06 '11 at 23:05
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    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
16

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

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

Willie Wong
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  • 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 06 '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 07 '11 at 10:31
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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 $$b = \exp \left( \frac{\ln a}{c} \right).$$

Fabian
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    @Fabian: Is it conventional to use "exp(...)" rather than "e^(...)"? I noticed that both you and picakhu (in the other answer) did this. – The Chaz 2.0 Apr 06 '11 at 15:26
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    @TheChaz: yes. See this. – Willie Wong Apr 06 '11 at 15:29
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    @The Chaz: excuse the pun, but it is a TeXnicality. – picakhu Apr 06 '11 at 15:30
  • @The Chaz: what is wrong with exp? that is how it reads on the button of my calculator... – Fabian Apr 06 '11 at 15:33
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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. – Ross Millikan Apr 06 '11 at 15:46
  • 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??? :) – The Chaz 2.0 Apr 06 '11 at 16:48
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Why we need exponential for everything? Answer could be simply as:

Log of $30$ with base "$b$" could be rewrite as $\log 30$ by $\log b$ regard to base $10$ resp. $\log 30$ is like $\log 3+ \log 10$ ( which is $1$ actually!) so, $\log 30=1.47712$

Now we need to find inverse $\log$ of ($1.47712 \times 0.3029$). or $10$ raised by that number which will come $75265.7$ approx.

TShiong
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Ready 2 Watch
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community May 10 '23 at 19:38
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In Excel: to quickly calculate the elusive $e$:

To calculate $e$ (the base of $\ln$): $e = x^{\frac1{\ln(x)}}$ Wherein: $x>1\text{ or }< 1\text{ but }x> 0$

Тyma Gaidash
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Jack
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