Logarithms: Finding the exponent given a base & an argument

Question

My question is this: In a logarithm, given a base and an argument, how do you derive the exponent (without counting, trial and error or educated guessing)?

e.g.

log2 (2048) = ?

If:

log() = e

I would like to know the algebra/formula for deriving the exponent(e) when you plug in a base(x) and an argument (y).

What I am seeking is not an algorithm but rather a formula, if there is one.

Thank you for reading

Answer 1

If $log_b(a)=x$ and you were asked to find x then this means you need to find some number x such that: $b^x=a$

In your example you are asked to find x in this: $log_2(2048)=x$

So you need some value for x such that: $2^x=2048$

x=11 satisfies this condition.

If you want to calculate $log_b(a)$ on a calculator that only has the $log_{10}$ function then you can use the change of base formula as follows:

$log_b(a)$=$\frac{log_{10}(a)}{log_{10}(b)}$

If you would like to calculate it by hand then, for your example, you would start with 2 and keep multiplying it by itself until you reached the answer. This would only be possible if the answer is an integer. So again for your example you would do:

2

2x2=4

4x2=8

8x2=16

16x2=32

…

1024x2=2048

And then count how many steps this took which should be 11 for your example.

If you want an algebraic formula then here is one that is valid for x>0:

$ln(x)=2\sum_{n=1}^{\infty}\frac{(\frac{x-1}{x+1})^{2n-1}}{2n-1}$

The above is an infinite expansion for log base e of x, typically written as $ln(x)$. So, if you want to calculate the log in some other base then you would first need to use the change of base formula to convert it to log base e.

Answer 2

So, by hand, with no calculator, logarithm tables or mechanical devices? Also, I understand that $2048$ and $2$ are just examples, and you would like a general method?

The simplest way is trial and error: Given base $b>1$ and argument $y>0$, try different values $x$ in order to find $b^x = y$. Do the exponentations by hand (by repeated multiplication).

Now of course you can be clever about what values $x$ you try. Surely if you tried some $x$ and got $b^x < y$, then you don't try that $x$ again (it will give the same answer again), nor do you try any smaller values of $x$ (they will give smaller $b^x$). So you increase your $x$ in order to increase $b^x$. Perhaps you increment $x$ by one, or perhaps more. Incrementing $x$ by one means you multiply $b^x$ by $b$; there are other methods, for example doubling $x$ means squaring $b^x$.

So perhaps you can find two values $x_1$ and $x_2$, such that $b^{x_1} < y < b^{x_2}$. This is called bracketing the solution: because $b^x$ is an increasing function of $x$, you know that the solution is somewhere between $x_1$ and $x_2$. The simplest thing you could now do is try some value $x_3$ strictly between $x_1$ and $x_2$ and try again: this way you can narrow down your bracket, as narrow as you want.

Now if the solution is actually an integer, this process will terminate when you find the integer. If not, you will have to use rational noninteger exponents $x=p/q$; either by taking $q$th roots by hand (not so nice but doable), or by observing that $b^{p/q}=y$ is equivalent to $b^p = y^q$ (probably nicer for hand calculations). But of course it could be that the solution is not even a rational number. Then the best you can do (by hand) is an approximation, or lower and upper bounds on where the solution is, to some accuracy that you are satisfied with. Repeated bracketing will do that for you.

There are other, more sophisticated methods but usually it is wise to start from understanding the simplest methods first.

Let's see how the bracketing might work with $b=2$ and $y=2048$, if we are not clever or lucky enough to know already the solution.

• Try $x=1$, so $b^x=2$. Oops, too small. Double the $x$ (and square the $b^x$).
• Try $x=2$, so $b^x=2^2=4$. Oops, too small. Double the $x$.
• Try $x=4$, so $b^x=4^2=16$. Oops, too small. Double the $x$.
• Try $x=8$, so $b^x=16^2=256$. Oops, too small. Double the $x$.
• Try $x=16$, so $b^x=256^2=65536$. Oops, too big. Try something between $8$ and $16$.
• Try $x=12=8+4$, so $b^x = 256\times 16 = 4096$. Oops, too big. Try something between $8$ and $12$.
• Try $x=10=8+2$, so $b^x = 256\times 4 = 1024$. Oops, too small. Try something between $10$ and $12$.
• Try $x=11=10+1$, so $b^x = 1024\times 2= 2048$. Success. We are done.

Note that at many points of the process you might have made "better" choices by educated guessing (e.g., after $x=12$ you might notice that $4096$ is not too badly off, only by a factor of two so you could try $x=11$ next). But since you specify "no educated guessing", you need a mechanical rule: halving the bracket each time is one, and gives you an algorithm that is guaranteed to finish if the result is an integer, and is guaranteed to give ever finer approximations otherwise.

---

Update (after edits on the question). Originally your question asked for "the method" that produces the result. Then it changed to "the algorithm/algebra/formula". But then the whole question turned upside down: now it is "not an algorithm but rather a formula". I am afraid that answering a question that moves goalposts is going to be difficult. It is not at all clear what kind of "formula" you're willing to accept.

At any rate, there is no fixed-size formula using just the four elementary arithmetic operations ($+-\times\div$) where you could plug in the two numbers and get an exact result. So something more will be needed, whether it is formula size (would a series expansion be allowed?), allowed elementary operations (would you have natural logarithm available?), method (can you do the calculation in steps and make choices on the way? That leads to an algorithm, and I don't understand what is wrong with that), or accuracy (for an approximation you could use, for example, a truncated series expansion). Second-guessing your thoughts (what is acceptable and what is not) may be difficult, to put it mildly.

At some point perhaps one needs to say:

Looks like the original problem ("the method that produces the result") is solved isn't it? As for "not an algorithm", that would be a different question, consider posting it separately.

Answer 3

There's no easily useable formula. There are quick algorithms though (continually square, therefore double your counter)

An example is $$\log_2(2^{2^{20}})$$ which is calculatable in just $21$ steps (assuming you can't see if off hand) .