Question

How do you solve `log_10 2`?

Answer 1

`log_10 2 ~~ 0.30103`

Explanation:

One somewhat cumbersome way of calculating `log_10 2` goes as follows:

First note that `2 < 10^1`, so `log_10 2 < 1` and we can write `color(blue)(0.)` as the start of the logarithm.

If we raise `2` to the `10`th power, then the effect on the logarithm is to multiply it by `10`.

Let's see what happens:

`2^10 = 1024 >= 1000 = 10^3`

So the first digit after the decimal point is `color(blue)(3)`

Divide `1024 / 10^3` to get `1.024`

To find the next digit of the logarithm, calculate `1.024^10` and compare it with powers of `10`:

`1.024^10 ~~ 1.26765060022822940149`

This is still less than `10^1`, so the next digit of the logarithm is `color(blue)(0)`

Then:

`1.26765060022822940149^10 ~~ 10.71508607186267320891 >= 10^1`

So the next digit of the logarithm is `color(blue)(1)`

Divide `10.71508607186267320891` by `10^1` to get:
`1.071508607186267320891`

Then:

`1.071508607186267320891^10 ~~ 1.99506311688075838379`

This is still less than `10^1`, so the next digit of the logarithm is `color(blue)(0)`

Then:

`1.99506311688075838379^10 ~~ 999.00209301438450246726`

That is very close to `1000=10^3`, so a very good approximation is to stop here with a digit `color(blue)(3)`, bearing in mind that it would actually be very slightly less.

Putting our digits together `log_10 2 ~~ 0.30103`