## Introduction to the Logarithm in Mathematics

The * logarithm* is the inverse operation of exponentiation.

Let’s define “log[b](b^x) = x” for non-zero real numbers ‘b’ and ‘x’. The image here shows how this is usually printed.

We retain the term “base” for logarithms, just as it was for exponents.

In the coin-toss experiment, 2^10 = 1024; so log[2](1024) = 10.

One trivial result is that “log[b](b) = 1” for all non-zero values of ‘b’, since “b = b^1”.

## ‘e’ is Euler’s Number and Napier’s Constant for Natural Logarithms

Say hello to our little friend ‘e’, a favourite of the famous mathematicians Euler and Napier.

The constant ‘e’ is approximately 2.71828…; it is a non-rational number that is the base for * natural logarithms*.

Mathematicians find ‘e’ and natural logarithms so useful that they reserve the notation “**ln**(x) = log[e](x)”.

As the image shows, “ln(e)” is defined as the “definite integral from 1 to ‘e’ of dx/x”. As well, since “ln(e) = log[e](e)”, 1 must be the value of that definite integral.

## The Reason to use a Logarithm to Convert an Exponent’s Base

At the outset, my problem was to solve for ‘x’ in the equation “2^m = 10^x”, where ‘m’ is a known positive integer and ‘x’ is a positive real.

Thanks to logarithms, this is equivalent to solving “log[2](2^m) = log[10](10^x)”.

## The General Rule for Converting Logarithms

The general rule for converting a logarithm from base ‘b’ to natural logarithms in base ‘e’ is “log[b](y) = ln(y)/ln(b)”.

We will use this rule in a later section.

## The General Rule for the Logarithm of an Exponential Term

In general, “log[b](x^m) = m*log[b](x).

Therefore “ln(2^m) = m*ln(2)”, which we will need in the next section.

## Deriving the Exponents’ Conversion from Base-2 to Base-10

Let’s step through the process for my original problem.

- “2^m = 10^x”, to solve for ‘x’.
- “m = log[2](2^m)” on the left side.
- “x = log[10](10^x)” on the right side.
- Apply “log[b](y) = ln(y)/ln(b)” to each side.
- “log[2](2^m) = ln(2^m)/ln(2)” on the left.
- “log[10](10^x) = ln(10^x)/ln(10)” on the right.
- Obviously “log[10](10^x) = x”, to be substituted soon…
- But “2^m = 10^x” in /#1/, so we substitute “2^m” for “10^x” in /#6/ as follows…
- “x = ln(2^m)/ln(10)” by lines /7/ and /8/.
- “x = m*ln(2)/ln(10)”, by the rule for the logarithm of an exponential.

A bit of work on a calculator gives the conversion factor, “ln(2)/ln(10) = 0.69314718055994530941723212145818 / 2.3025850929940456840179914546844 = 0.30102999566398119521373889472446” to a large degree of accuracy.

I’m likely to use * 0.301* as the conversion factor.

Let’s check my original problem, which was “2^10 = 1024 = 10^x”. My conversion is “x = 10 * 0.301 = 3.01”.

On the calculator, “10^3.01 = * 1023.29*29922807541309662751748199″, which is fairly close to 1024.

John says

Re OP:

same idea, easier for me to understand:

1. “b^m = a^x”, to solve for ‘x’.

2. “log[a}(b^m) = x”. take log[a] of both sides.

3. “m * log[a}(b) = x”. pull the m out of the log

4. “m * ln(b)/ln(a) = x”. replace with natural logs

Re Chuck:

from what you described:

b^m = a^x

b/a = c

c * m =? x

c should be 0.301 from OP’s calculations rather than 5

also log[2]( 10^15 ) = 49.829 not 75

in other words:

10^14 = 2^46.507

10^15 = 2^49.829

Chuck Crosthwait says

I’m pretty sure I found an easier formula for converting an exponent series from base10 to base2. I’ll try to express it here with a known value.

10^14=2^x

Using the formula above you’d use

x=14*ln(10)/ln(2)

and arrive at 70.

If you use some funky cross multiplication I haven’t seen anywhere you could do the following instead:

10^14=2^x

10/2=5

5*14=70

10^14=2^70

I’d like to preface with the fact that I did not locate this formula anywhere after searching for ~1hr. I tried to remember the general formula for finding exponent X in a known system. I have only tried this converting from base10 to base2 and back and tried with other known values, e.g.:

10^15=2^x

10/2=5

5*15=75

10^15=2^75

and this is borne out in the formula

x = m*ln(b)/ln(a)

for finding

b^m = a^x

so it works. Did I stumble upon a formula I haven’t found before or is this something new and just too simple to have been considered before?

Please do reply, I really want to know if there’s some unknown mathematical principle I needed to know or knew and forgot.