Quick way to estimate powers of $9$?

Question

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 $(+,-,\times,/)$. What would be the fastest or most efficient way estimate $2^n$ or $9^n$

Answer 1

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.

Answer 2

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 $9^{12} \approx 2.8243 \cdot 10^{11}$.