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Is there any way to calculate/approximate values like $$(\frac{125}{250})^{0.66}$$ using only a pen, paper and the mind?

(Above expression being just an example, the numbers may vary and not be easy to approximate)

I've already tried the expansion method but I seek a $\mathbf {faster}$ $\mathbf{method}$ that can be used in an exam. Nothing else comes to my mind, nor could I find any other method on the internet.

Thank you in advance.

CapHim YT
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    You can calculate logs, antilogs, sins etc using the expansion formula. Theoretically possible. Practically difficult to do. – Aatmaj Sep 15 '23 at 12:36
  • @Aatmaj Thank you for the suggestion! Unfortunately, I'm looking for a method that can be used in an exam and is not too time consuming. – CapHim YT Sep 15 '23 at 12:39
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Sep 15 '23 at 12:57
  • Hi @Shaun I'm not sure what else I can add to my edit regarding this question. I was just looking for someone who knows a method to approximate such values as I'm not able to find one myself. Thank you for the tip though! – CapHim YT Sep 15 '23 at 13:08
  • Was it intentional to have the fraction reduce to $\frac12$ in the example? – jackson Sep 15 '23 at 13:10
  • @jackson No, it was just an example. They can be any random numbers like $\frac{19}{17}$ . – CapHim YT Sep 15 '23 at 13:13
  • I have given some discussion in this question and this question. There are many tricks that depend on knowing facts. How much work it is depends on how accurate you want to be. – Ross Millikan Sep 15 '23 at 13:56
  • @RossMillikan Can you please explain how these methods would help me in calculating values like $[\frac{19}{17}]^{0.66}$ ? I'm having trouble figuring out how to use approximations in log to calculate an exponential value. – CapHim YT Sep 15 '23 at 14:24
  • See Calculating logs and fractional exponents by hand – Jam Sep 15 '23 at 15:45
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    I'm looking for a method that can be used in an exam and is not too time consuming --- Just what kind of exam would you need this for?? – Dave L. Renfro Sep 15 '23 at 17:19
  • For that the first approximation comes from $(1+x)^k \approx 1+kx$ for $x,xk \ll 1$, so it would be $1+\frac 2{17}\cdot 0.66$, easy to do by hand – Ross Millikan Sep 15 '23 at 18:51
  • Not all of them are as easy. The test maker should make sure they are. The $.66$ looks very close to $\frac 23$ to me, so you can square and take a cube root for that part. – Ross Millikan Sep 15 '23 at 20:02
  • @RossMillikan what should I do if the condition x << 1 isn't fulfilled? For instance : $[\frac{127}{13}]^{2.66}$ . – CapHim YT Sep 15 '23 at 20:04
  • @RossMillikan Though I don't want to take chances but I see that you have a point. Thank you for your help! – CapHim YT Sep 15 '23 at 20:08

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I don't have a good solution for your example with a decimal exponent, but in the title you asked about fractional powers. You can approximate these pretty easily using Newton's method. Modifying your example to $(\frac{125}{250})^\frac{2}{3}$, we can use $x_0=\frac{27}{64}$ to get $\frac{27}{64}^{\frac{2}{3}}+(\frac{125}{250}-\frac{27}{64})\frac{2}{3}(\frac{27}{64})^{-\frac{1}{3}}$. Simplifying this gives $\frac{9}{16}+\frac{5}{72}=\frac{91}{144}\approx0.632$, which is pretty close to the actual value, $0.630$. This is pretty feasible if the numerator and denominator are single digits, but once you start dealing with more complex fractions, it starts to become very difficult to find a good $x_0$, unless you're lucky enough to be able to use $x_0=1$.

We can convert any decimal exponent into a fractional exponent. For example: $(\frac{125}{250})^{0.66}=(\frac{125}{250})^\frac{66}{100}$. Unfortunately, I don't know any power of $\frac{66}{100}$ off the top of my head, aside from the trivial $x_0=1$, so we'll use that. We get $1+(\frac{125}{250}-1)\frac{66}{100}(1)^{-\frac{34}{100}}=\frac{66}{100}=0.66$, which is sort of close to $0.633$.

If we always use $x_0=1$, we can get a formula of $x^n=1+nx-n$, which is easy to memorize, and works reasonably well for values from $n<x<\frac{1}{n}$

Sebastian Nuxoll
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    Thank you for helping me with this! Can you just explain what $ \mathit x _0$ is defined as in the first example and the formula if we don't use $ \mathit x _0$ =1? – CapHim YT Sep 16 '23 at 15:41
  • We want $x_0$ to be as close to the base of the exponent as possible, while still having and exponent that's easy to calculate. In my first example, we chose $\frac{27}{64}$, because its pretty close to $\frac{125}{250}$, and because we already know the exponent: $(\frac{27}{64})^\frac{2}{3}=\frac{9}{16}$ – Sebastian Nuxoll Sep 17 '23 at 20:32
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    We then use the formula $f(x_0)+(x-x_0)f'(x_0)$ – Sebastian Nuxoll Sep 17 '23 at 20:35
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    Thank you for the clarification! – CapHim YT Sep 18 '23 at 02:30