How to find out the value of numbers having fractional powers manually without using logarithms and calculators??
For example : $2^{1.6}, 3^{2.1}, 5^{3.22}$ etc, I know we can find out the value using logarithms. But I want to know how to find out the value manually. Calculators can easily find out the values. That means some logic should be built to find out the value. Please help me to know this?

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    $\begingroup$ Powers are defined by the exponential function: $a^x=e^{x\ln(a)}$ for $a>0$, $x$ real. For rational values of $x$ you can also use $a^{\frac{p}{q}}=\sqrt[q]{a^p}$. $\endgroup$ – J.R. Feb 5 '14 at 9:37

What D.L. gave as an answer is a good way which can be slightly improved for faster convergence of the series.

Suppose that we write $x=\lfloor x\rfloor+y$. So we could use $$a^x=a^{\lfloor x\rfloor }\sum_{n=0}^{\infty} \frac{(ln(a).y)^n}{n!}$$ This could be for $y<\frac{1}{2}$. Otherwise we should use $x=\lceil x\rceil-y$ and use $$a^x=a^{\lceil x\rceil}\sum_{n=0}^{\infty} \frac{(-ln(a).y)^n}{n!}$$ So, you have integer powers of $a$ and a very few logarithms to remember.


You can use the formula $a^x=\sum_{n=0}^{\infty} \frac{(\ln(a).x)^n}{n!}$, and calculate the first terms. You will only have to know $\ln(a)$. But it is maybe what you call logarithms method ?

For example, $2^{1.6}= 3,03$ and $1+ \ln(2).1,6 + \frac{(\ln(2).1,6)^2}2=3,09 $ with the approximation $\ln(2)=0,8$.


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