# Quick way to estimate powers of $9$?

I was trying to estimate how much is $$9^{12}$$, and I came with my own formula that does the work with powers of $$10$$, that is : $$9^n\approx|10^n-n \cdot 10^{n-1}|$$ but the larger the $$n$$ (integer) larger is the error.

Is there a way to simpler way to estimate at least the power of $$10$$ of $$9^n$$ for large $$n$$?

Same question for powers of $$2$$.

Edit: to make it more specific, suppose that you only have a calculator with the four basic operations (+,-,*,%). What would be the fastest or most efficient way estimate 2^n or 9^n

• You could use logarithms or the binomial theorem to get estimates (your estimate, without the absolute value, would be a binomial estimate) – Mark Bennet Aug 11 '19 at 12:52
• To approximate $9^n$ with powers of $2$ you can use the fact that $9 = 2^3 + 1$ and do a binomial expansion to however many terms you like. – 0XLR Aug 11 '19 at 12:54
• The number of digits of $9^n$ is $\lfloor n\log_{10}9\rfloor$. – mathcounterexamples.net Aug 11 '19 at 12:55
• What do you mean by estimate $9^12$? That's already an integer. – Allawonder Aug 11 '19 at 12:55
• $9^n=10^{n\log_{10}(9)}\approx10^{0.954n}$ – Simply Beautiful Art Aug 11 '19 at 12:56

For powers of $$2$$ the approximation $$2^{10} = 1024 \approx 1000 = 10^3$$ is your friend. Then $$2^{20}$$ is about one million, and so on.

For, say, $$2^{35}$$, use the fact that $$2^5 = 32$$ to get the approximation $$32,000,000,000$$.

For powers of $$9$$, use what @SimplyBeautifulArt has in his comment:

$$9^n=10^{n\log_{10}(9)}\approx10^{0.954n} .$$

That means $$9^n$$ has about $$0.95n$$ base $$10$$ digits when $$n$$ is large.

• Wow nice! $(+1)$ – Mr Pie Aug 11 '19 at 13:08

Following on from Simply Beautiful Art's comment,

$$\log_{10} 9 = \frac{\ln 9}{\ln 10} = \frac{\ln(9/e^2) + 2}{\ln(10/e^2) +2}$$

Let $$a=9/e^2 - 1$$ and $$b=10/e^2-1$$. Then we can use the Taylor series of $$\ln(1+x)$$:

$$\ln 9 = 2 + a - \frac{a^2}{2} + \frac{a^3}{3} - o(a^4)$$

$$\ln 10 = 2 + b - \frac{b^2}{2} + \frac{b^3}{3} - o(b^4)$$

Using this estimate, $$\ln 9 \approx 2.1977$$ and $$\ln 10 \approx 2.3056$$, so $$10^{12 \cdot 2.1977 / 2.3056} \approx 2.7434 \cdot 10^{11}$$, compared to $$9^{12} \approx 2.8243 \cdot 10^{11}$$.