1
$\begingroup$

I'm a Year 12 Specialist Mathematics student currently studying logarithms. In class we constantly use the calculators for logarithms. does any-body have a simple way of doing logarithms without a calculator or log-book?

For instance a simple one like $$\log_{3}22=x$$
How would I solve for $x$?

  • 1
    Unless you like infinite series expansions then no. – 2017-02-02
  • 1
    Depends on what you mean by "doing". I would hate to have to compute $\log(22.36523)$ to $7$ decimal places by hand. But I could give you a pretty good approximation of $\log(1.0002)$ (especially if you want a natural logarithm). – 2017-02-02
  • 0
    Are you asking about [simple cases like $\log_8128 = \frac 73$](http://math.stackexchange.com/questions/1653374/easy-way-to-compute-logarithms-without-a-calculator)? – 2017-02-02
  • 0
    $\log_3(22)$ does not have a "simple" expression. It is a transcendental number. – 2017-02-02
  • 0
    Of course you could write it as $\log_3(2) + \log_3(11)$. But I wouldn't call that simple. – 2017-02-02
  • 0
    To how many decimal places? $3^2 = 9$ and $3^3 =27$ so $2< \log_3 22 <3$. That's usually as far as I want to go, but I can do $3^{2.5} = 9*\sqrt {3} $ and I can approximate $\sqrt 3 $ by hand. So this is tedious and inefficient but it *can* be done. I just don't think it *should* be done. – 2017-02-02
  • 0
    Have you seen people in 2017 using a log-book ? – 2017-02-02
  • 0
    No, it is not a '*simple*' general way to calculate the logarithm. – 2017-02-02
  • 0
    <$x$?>> You have *already* solved for $x$. What you mean is "how would I obtain a numerical approximation to the value of $x$" – 2017-02-02

1 Answers 1

1

In general there is no way to compute logarithms exactly without either using a computing device or hours of pen and paper calculations.

You case is further complicated by the fact that the base of the logarithm is not $e$. In other words, you example is not a "simple one". However, you can do something to simplify it:

$$\log_{3}22=\log_{3}(27-5)=3+\log_{3} \left(1-\frac{5}{27} \right)$$

(Remember, $27=3^3$). Because for $|t| \to 0$ we have $\log(1+t) \approx t$, we can conclude that $x \approx 3$, but $x<3$. The numerical value is about $2.8135880922155955$.

If you want, you can use the infinite series for $\log(1+t)$, however, you need to remember that in your case the change of base is necessary:

$$\log_{3} \left(1-\frac{5}{27} \right) = \frac{1}{\log 3} \log \left(1-\frac{5}{27} \right)=-\frac{1}{\log 3} \sum_{k=1}^\infty \frac{5^k}{k 27^k}$$

So, even to find an approximation by truncating the series for some finite $N$ (i.e. $\sum_{k=1}^N \frac{5^k}{k 27^k}$), you first need to compute $\log 3$ with high precision.

There are many ways to compute approximate values of logarithms (and calculators of course use some of them too), but they all involve long and difficult calculations.

I highly recommend the book Analysis by Its History , I think in the first chapter they have a long discussion about how people were computing logarithms in $18-19$th centuries, and how it lead to the development of modern methods.

Logarithms were extremely important for navigation at the time, so every ship had a logarithm table. And there were few people who could create these tables.