Recently I start realising the decline of my math instinct. The other day, it took me a few minutes to remember the Gaussian sum and more minutes for the very basic Trigonometry. The only thing about calculation remains sharp in my mind is probably the Chinese multiplication table. I’m having a hard time to admit that it would be hard for me to pass a high school entrance math tests. So here I am, making a little bit effort to pick up some old or new math tips and keep my brain running.

This chapter is about mental calculation. And it started with me came across this old website(NOTE: the URL starts with http://). For me it worth the risk of stepping into the old security protocol in exchange of a good time with $2^n$, $e^n$ and $\pi^n$.

Calculate $e^n$ (Source)

As a refresher, the Euler’s number (AKA Napier’s constant) $e \approx 2.718281828459045$.

The mental work of figuring out $e^x$ is by rounding it to the power of 10, so $e^x = 10^y$. Following more math rules, there is $$ log_{10}(e^x) = y \\
x\cdot log_{10}(e) = y \\
=> x \times 0.4342944… = y \\
=> x \times (43 \cdot 101)⁄10,000 = y $$

For example, calculating $e^{16}$ turns into:
1) multiplying by 43
$16 \times 4$ add $16 \times 3$ \rightarrow $64$ add $4$ with $8$ -> $688$

2) multiplying by 101
2 digit numbers multiplying by 101 is just repeat the 2 digit number itself.
More digit numbers will split the last two digits out and add up the split numbers to fill the middle positions. Think about $10^n \times 101$ for the total number of digits of the answer, so $XXX\times 101$ will have no more than 6 digits and $XXXX\times 101$ will have no more than 7 digits. And for both of those cases there are 2 positions in the middle to fill. e.g. $688 \times 101 \rightarrow 6 \square\square 88$ combined with $6 + 88 = 94 \rightarrow 69488$

3) dividing by 10,000
$69488 \rightarrow 6.9488$

4) adjusting for approximation
DROP the last (rightmost) digit of the answer, no rounding. $6.9488 \rightarrow 6.948$ thus $e^{16} \approx 10^{6.948} \approx 8.9 \times 10^6$

Calculate $\pi^n$ (Source)

Similar as above, solving the power by using the power of 10, $$ x\cdot log_{10}(\pi) = y \\\
x \times 0.49714987… = y \\
=> x \times 0.5 = y $$

For example, calculating $\pi^{16}$ turns into:
1) building the subtraction problem
NUMBER 1: multiply the power by 500, so $16 \times 1000 / 2 = 8000$
NUMBER 2: multiply the power by 3, $16 \times 3 = 48$

2) subtraction
Subtract the smaller number from the larger one. This gets $8000 - 48 = 7952$

3) adjusting for approximation
Take the number just subtracted, and drop the last digit, no rounding. Then divide the remaining digits by 2. If that ends in a .5, drop the .5 as well. So this happens on the NUMBER2
$48 \rightarrow 4 \rightarrow 2$.
Add the above result to the subtraction result, $2 + 7952 = 7954$.

4) dividing by 1000 for the y value
$7954 \rightarrow 7.954$
And $\pi^{16} \approx 10^{7.954} \approx 9 \times 10^7$

Calculate $2^n$ (Source)

Here it starts a little differently – memorize $2^0$ to $2^{10}$. And they are
$$ 2^0 = 1 \qquad 2^1 = 2 \qquad 2^2 = 4 \qquad 2^3 = 8 \qquad 2^4 = 16 \qquad 2^5 = 32 \\
2^6 = 64 \qquad 2^7 = 128 \qquad 2^8 = 256 \qquad 2^9 = 512 \qquad 2^{10} = 1024 $$ Since $2^{10}$ is close to $1000$, here takes $2^{10} \approx 10^3$.

For example, calculating $2^{16}$ turns into:
1) breaking the power
$2^{16} = 2^{10} \cdot 2^6$
With memory of the simple power of 2,
$2^{10} \cdot 2^6 = 2^{10} \cdot 64$

2) calculating the multiple 10
Replacing $2^10$ with $10^3$. In this case it becomes $64 \times 10^3 = 6.4 \times 10^4$ And this can the estimated result.

3) adjusting for approximation
The difference between $2^{10}$ and $10^3$ is $2.4\%$. For every $2^{10}$ approximation, multiply the percentage difference to get a more accurate answer.
$2^{16} \rightarrow 6.4 \times 10^4$ uses one $10^3$, the percentage difference is $2.4\% \times 1$.
To take the difference into account the result becomes $64 \times (1+2.4\%) \times 10^3 = (64 + 1.536) \times 10^3 = 6.5536 \times 10^4$

Calculate $10^{n.abc}$

The calculation starts by memorizing two basic logarithms: $$ log(2) = 0.30103 \qquad log(3) \approx = 0.48 \\
2 = 10^{0.30103} \qquad 3 = 10^{0.48} $$ And since $2^{10} / 10^3 = 1.024$, with above it’s not complicated to get $$ (10^{0.30103})^{10} / 10^3 = 1.024 \\
10^{10 \cdot 0.30103 - 3} \approx 10^{0.01} = 1.024 \\
10^{0.02} = 1.05 $$

For example, estimating $10^{7.954}$ turns into: 1) splitting the decimal power
$10^{7.954} = 10^{0.954} \times 10^7$

2) calculating number of $10^{0.3}$
$10^{0.954} \approx 10^{0.48 \cdot 2} = 3^2$

3) combining the result
$3^2 \times 10^7 = 9 \times 10^7$

That’s it. 🎉 🎉

More reading:
Calculating Base 10 Logarithms
How can I calculate 10 to the power of a decimal number