how to calculate the value $0.9^{31}$

Question

How to calculate the value of a decimal raised to a big power without a calculator. I have been using the formula $1-m x+ (m(m-1)/2!) x^2$ but this does not give the correct answer sometimes. Can someone approximate $0.9^{31}$?

Answer 1

Let $0.9^{31}=x$.

Then, $\log x=31 \cdot\log 0.9=31 \cdot (2\log3-1) \approx 31 \cdot (2 \cdot 0.48 -1)=-1.24 $

Thus, $ x \approx \frac{1}{10 \cdot 10^{0.24}}\approx \frac{0.1}{\sqrt[4]{10}}\approx 0.056$

Answer 2

If you wanted to calculate it to arbitrary precision you could use binary exponentiation. Personally this doesn't seem very fun to me but it would be a reliable if tedious way to get, say, $10$ digits of accuracy if that's what you wanted. Here's a calculation with $2$ digits of accuracy:

$$0.9^2 = 0.81$$ $$0.9^4 = 0.81^2 \approx 0.66$$ $$0.9^8 \approx 0.66^2 \approx 0.44$$ $$0.9^{16} \approx 0.44^2 \approx 0.19$$ $$0.9^{32} \approx 0.19^2 \approx 0.036$$

This gives $0.9^{31} \approx \frac{0.9^{32}}{0.9} \approx \boxed{0.04}$. The true answer turns out to be $0.9^{31} \approx 0.038 \dots$ so this is pretty good.

For a sort of silly alternative, write it as $\left( 1 - \frac{1}{10} \right)^{31}$ and use the fact that we know $\left( 1 - \frac{x}{n} \right)^n \approx e^{-x}$. This gives that the answer is approximately $e^{-3.1}$. This is not very useful unless you happen to know the values of some natural logarithms by heart, but if you do, you can use, for example, that $\log 2 \approx 0.69 \dots$ to write

$$e^{-3.1} \approx e^{-5 \log 2 + 0.35} = \frac{e^{0.35}}{32}.$$

Coincidentally it happens that $0.35$ is approximately $\frac{\log 2}{2}$, which altogether gives

$$e^{-3.1} \approx \frac{\sqrt{2}}{32}$$

and if you know that $\sqrt{2} \approx 1.4 \dots $ then dividing by $32$ gives

$$e^{-3.1} \approx \boxed{ 0.044 \dots }$$

so this actually does worse than binary exponentiation but we were admittedly quite sloppy at several points. However, this approach has the benefit of generalizing better to larger numbers; if we wanted to estimate, say, $0.99^{311}$ this estimate would work better than it did here because we'd be getting a closer approximation to the exponential.

This second calculation was ad hoc as presented but it can be systematized; if you want to estimate $e^x$ you do it by estimating some logarithms and subtracting those logarithms from $x$, and you try to get $x$ small enough that other estimates such as the Taylor series estimate become accurate. In turn you can estimate logarithms by hand using Taylor series in a clever way; see this math.SE answer for some details.