Question

Given that log 5.0 = 0.6990 and log 5.1 = 0.7076, find to the nearest hundredth a value of x for which log x = 0.7060. Can anyone solve this and provide an explanation? Thanks!

Answer 1

5.08

Explanation:

There's a few ways to do this, but either way, we will use the first order Taylor/Maclaurin expansion for `ln(1+x)`:
`ln(1+x) = x + O(x^2)`

Let's define (with `log(x) = log_10(x)`) the quantity `l` via
`log(5+chi) = l`
We know two cases:
`chi = 0 implies l_0 = log(5)`
`chi = 0.1 implies l_1 = log(5.1)`

We will assume this is first order. We can use two log rules in order to simplify for `l`, then use Taylor expansion:
`l = log(5+chi) = log(5 (1 + chi/5)) = log(5) + log(1 + chi/5)`
`l = l_0 + log_10(e) * ln(1+chi/5) approx l_0 + log(e)*chi/5`

We could use our calculator to get a value for `log(e)`, but we could also use the second condition given to us. Using our definition for `l_1`, we can set `chi = 0.1` and find that
`l_1 = l_0 + log(e) * 0.02 implies log(e) approx 0.43`

Solving for `chi` in terms of `l`,
`l = l_0 + 0.43/5 chi implies chi = (l-l_0)/(0.086)`

Plugging in the numbers given,
`chi = (0.007)/0.086 = 7/86 approx 0.0813...`

Therefore, we know the answer to 2 significant digits is 5.08.

With the linearity condition, there's actually a bit of a quicker way to reach it. Linear functions have a nice condition: if an input is `y` of the way between a and b, its output is `y` of the way between the outputs of a and b. This means that we can find this `y` and then figure out the answer.

`x = 5 + 0.1y`
`y * l_0 + (1-y) * l_1 = l_text(find)`
`y = (l_f - l_1) / (l_0 - l_1) = (0.0086)/(0.007) = 86/70`

This agrees exactly with the above.

If we had just thrown the original into a calculator, we would have found that
`log(x) = 0.7060 implies x = 5.08159...`
showing the power of this method and that the error only appears two digits down, which is that second order term we originally mentioned.