# How exactly do logarithms work?

1. Oct 4, 2009

I was just wondering what's the "magic" behind logarithms. Just took them in school, and I get the part that they're just the reverse of exponents but it still feels sort of like a mystery every time I solve a logarithm using a calculator. Just like how addition is the slow way of multiplication: $$6\times3 = 6 + 6 + 6$$ and multiplication is the slow way of raising something to some power: $$6^3 = 6\times6\times6$$ so surely there must some long/slow way to do logarithms that was in use before they were discovered/invented? $$(\log _{2}^{32} = 5) = ???$$

Hopefully someone can shed some light on this, it's sort of bugging me that I have no idea what goes on behind the scenes, Thanks ^^

Last edited: Oct 4, 2009
2. Oct 4, 2009

### symbolipoint

The more reliable characterization is Logarithms are exponents. Logarithms are not reverses of exponents; logarithmic functions are the inverses of exponential functions.

One first learns the rules of exponents with a common base. Derivations and exercises help to fix these rules. Later, when one studies functions, one reaches exponential functions, does some exercises, and then examines the idea of the inverse of an exponential function. Finally, one is shown the rules of logarithms, which are the same as the rules of exponents because logarithms ARE exponents. The notation appears different for logarithms than for exponents. Because this has a different appearance, students struggle with initial confusion.

Learn how to transcribe between exponential forms and logarithmic forms of equations; doing so is often very helpful:

b^x = y <------> LOG_b (y) = x [the logarithm base b of y equals x]

y = b^x is the inverse of y = LOG_b (x)

3. Oct 5, 2009

Thanks for the reply but I think I might have worded my question wrongly... I know how to use logs and I know they're the inverse of exponential functions but was wondering how exactly would one go about computing logarithms without using the log function. Like if I wanted to figure out x in $$4^x = 1024$$ without using logs and without guessing how would I do it? I guess I'm trying to find out how people solved that for x before logarithms were discovered.

The only thing close to what I was looking for is Borchardt's Algorithm: $$ln(x) \approx \frac{6\times(x - 1)}{x + 1 + 4x^{0.5}}$$
I was looking for something like that, something that allows us to solve problems that include something raised to x without logs.

Thanks again symbolipoint but wasn't what I was looking for :)

//EDIT: What I meant by "I have no idea what goes on behind the scenes" is that I have no idea how the calculator calculates logs, surely it doesn't guess and it doesn't have a list of values ready, it must have computed my log by using a combination of more elementary processes, and that's sort of what I want to know.

Last edited: Oct 5, 2009
4. Oct 5, 2009

### HallsofIvy

You could, for example, use the Taylor's series:
$$log(1+x)= \sum_{n= 1}^\infty \frac{x^n}{n}= x+ \frac{x^2}{2}+ \frac{x^3}{3}+ \cdot\cdot\cdot$$

5. Oct 5, 2009

### HallsofIvy

You could, for example, use the Taylor's series:
$$log(1+ x)= \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}= x- \frac{x^2}{2}+ \frac{x^3}{3}- \frac{x^4}{4}+ \cdot\cdot\cdot$$

6. Oct 5, 2009

### symbolipoint

HallsofIvy has given examples of Taylor Series formulas, which are developed in the study of Calculus, usually the Second Semester course of College Calculus. Look at this as a way of being motivated to keep studying Math courses at least through Calculus 2.