Question

What is the value of log 43?

Answer 1

`log 43 ~~ 1.63`

Explanation:

`43` is a prime number, so `log 43` cannot be expressed in terms of logarithms of smaller numbers.

Using a calculator, we find:

`log 43 ~~ 1.63346845558`

How could we find it by hand?

One somewhat arduous method goes as follow:

Note that `10^1 <= 43 < 10^2`, so the portion of the logarithm before the decimal point must be `color(red)(1)`

Dividing `43` by `10` (and thus subtracting `1`) from the logarithm, we get `4.3`.

If we raise this to the tenth power, then its logarithm will be multiplied by `10`, i.e. shifted one place to the left.

We find:

`4.3^10 = 2161148.2313284249`

So:

`10^6 <= 4.3^10 < 10^7`

So the next digit of the logarithm is `color(red)(6)`

Divide `2161148.2313284249` by `10^6` to subtract `6` from its logarithm and raise to the tenth power to shift the logarithm another place to the left:

`2.1611482313284249^10 ~~ 2222.519`

Then:

`10^3 <= 2222.519 < 10^4`

so the next digit is `color(red)(3)`

Keep on going for as many digits as you want.

Thus far we have found:

`log 43 ~~ 1.63`

Answer 2

`log 43 ~~ 1.633`

Explanation:

Suppose you know that:

`log 2 ~~ 0.30103`

`log 3 ~~ 0.47712`

Then note that:

`43 = 129/3 ~~ 128/3 = 2^7/3`

So

`log 43 ~~ log(2^7/3) = 7 log 2 - log 3 ~~ 7*0.30103-0.47712 = 1.63009`

We know that the error is approximately:

`log (129/128) = log 1.0078125 = (ln 1.0078125) / (ln 10) ~~ 0.0078/2.3 = 0.0034`

So we can confidently give the approximation:

`log 43 ~~ 1.633`

A calculator tells me:

`log 43 ~~ 1.63346845558`