Calculating power without using a calculator, for example $1.05^{10}$ How to find (or estimate) $1.0003^{365}$ without using a calculator? Do we have any fast algorithm for cases where base is slightly more than one? Say up to $1.1$ with tick $0.05$.
 A: You could try the binomial expansion of $(1+0.05)^{10}$ and stop calculating terms after they become small enough to not affect your required degree of accuracy
A: $$(1+x)^n\approx 1+nx\quad(x\ll1)$$

$$^*(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...$$
A: Well binomial theorem with some basic algebraic manipulations,I find it easier to play with integers than decimals.
$$1.05^{10}=\frac{105^{10}}{100^{10}}=\frac{5^{10}\cdot21^{10}}{5^{10}\cdot20^{10}}=\frac{(20+1)^{10}}{20^{10}}=1+\frac{10\cdot20^9+45\cdot 20^8+{10\choose 3}20^7+\cdots+1}{20^{10}}=1+1/2+\frac{45}{20^2}+\frac{120}{20^3}+\cdots+\frac{1}{20^{10}}=1.5+0.1125+0.015+\cdots+\frac{1}{20^{10}}\approx 1.6275$$
A: First, learn how to work with base-10 logs in your head, including converting them back and forth. Here's a simple video that will teach you how to do this: https://www.youtube.com/watch?v=V5rgTPu8JcE
Next, realize that $e$ can be used to approximate such problems. A good understanding of $e$ can be found here: http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/
Finally, learn to take equations in the form of $e^{x}$ and convert them into equivalent problems of the form $10^{y}$: http://headinside.blogspot.com/2014/03/calculate-powers-of-e-in-your-head.html

To actually work through the problem, multiply the interest (0.05 in your example) times the power (10 in your example), and start by raising $e$ to that power. 0.05 × 10 = 0.50, so we're looking at roughly $e^{0.50}$.
We're going to convert this into the form $10^{y}$, but the conversion method works better with integers, so we should the equivalent of $e^{50}$, work out $y$ for $10^{y}$, and then move the decimal place of $y$ two spaces to the left to compensate.
Using the method from the calculate powers of e tutorial above, $e^{50}$≈$10^{21.715}$, so $e^{0.50}$≈$10^{0.21715}$. Now, we just have to work out what the base-10 antilogarithm for 0.21715.
As explained in the youtube link above, a 0.041 difference between logarithms corresponds to a roughly 10% change in the equivalent decimal. 0.217 + 0.041 + 0.041 = 0.299, which is quite close to the base-10 log of 2. Two differences of 0.041 mean two differences of 10%. In short, the answer should be roughly 20% less than 2, which is 1.6.
OK, this probably isn't a quick a method as you may like, but it does return a decent estimate.
A: You can use successive squaring.
$$x^{10} = x^2(x^4)^2 = x^2((x^2)^2)^2$$
so first calculate $x^2$, then save it for later, calculate square two more times and then multiply with the $x^2$ you saved. This will make smaller the number of multiplications you need to do. From 9 down to 4. Will be much more impressive for large powers and powers that are a multiple of two. For example $x^{1024}$ will require no more than 10 multiplications!
