We can write $3^{0.6}$ as $3^\frac6{10}$ and can be simplified to $3^\frac3{5}$, yet it is tedious to calculate by hand, as well as $3^e$ and $3^{\pi}$? How can I do this without calculator? Like how can such things be calculated by hand?
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3Can you clarify what you mean by "simplified"? Do you mean to compute the numerical value (e.g. $3^e = 19.812\cdots$), or do you want an analytic expression of some form? – Trebor Oct 03 '24 at 06:47
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2Usually, you compute $a^b$ as $e^{b\ln a}$, and $e^t$ is computed by $$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots$$ – ultralegend5385 Oct 03 '24 at 06:56
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Mh, evaluating a tenth (or a fifth) root, is by no means easy. How do you perform that? – Oct 03 '24 at 07:59
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@Trebor Sorry, I should have clarified enough, I meant that 3^6/10 can be written as 3^3/5. – Jahraa Kabir Rose Oct 03 '24 at 08:43
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@YvesDaoust Yes, it's not easy. I am editing the question. Thanks for pointing out. – Jahraa Kabir Rose Oct 03 '24 at 08:43
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1Is the use of log tables "by hand"? If so, use $\log a^b=b\log a$. – J.G. Oct 03 '24 at 09:43
2 Answers
In your example since $0.6$ is a rational number, you can express it as $\frac{3}{10}$. But $e$ and $\pi$ are examples of irrational numbers so you cannot express them as a fraction. The best approach is to use the series expansion of $a^b = e^{b\ln a}$ as explained in the comments to compute an estimate of the numerical value to a desired degree of accuracy.
But if you just want to express it in a form where the exponent is some fraction as in your example then, although you cannot exactly represent $e$ or $\pi$ as a fraction due to their irrationality, you can always find rational approximations of irrational to any desired degree of accuracy using the continued fractions. Then you can express it is
$$ 3^{\pi} = 3^{\frac{a_n}{b_n}} + \text{an error term depending on $n$} $$
where $\frac{a_n}{b_n}$ is a convergent, $n = 1,2,3,\ldots$. The bigger is $n$ the better is your the estimate $3^{\frac{a_n}{b_n}}$ and the smaller is the error is error term. For example the first few convergents of the continued fraction of $\pi$ are
$$ 3, \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \frac{103993}{33102},\ldots $$
Most well known constants have well known continued fractions but in general finding the continued fraction of an arbitrary irrational number involves numerically estimating the number to several decimal places so at end of the day, this method is not any more efficient then directly estimating the value using the series expansion of $e^t$.
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If you are skilled in written computation, you can form a table of successive square roots, starting from $3$ (i.e. exponents $\frac12,\frac14,\frac18\cdots$). This can be done using the direct method for square roots (https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Decimal_(base_10)), or by Newton's iterations.
Now write the power as a fractional binary number and compute the product of the corresponding powers.
$$\begin{align}1&\to 3\\\tfrac12&\to1.7321\\\tfrac14&\to1.3161\\\tfrac18&\to1.1472\\\tfrac1{16}&\to1.0711\\\tfrac1{32}&\to1.0349\\&\cdots\end{align}$$
Note that the next roots are obtained by just halving the fractional part.
$$\pi=11.0010010000\to 3^\pi\approx3^3\cdot1.1472\cdot1.0173=31.5103$$
(Only the first decimal is correct :-( )
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Could you please explain in more detail? Like can you please explain why we should use binary system? And I didn't understand the last line calculation. I am having some problems grasping this. – Jahraa Kabir Rose Oct 03 '24 at 09:00
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Binary makes the computation simpler as the roots are square and you need to compute one of each only. Using the decimal base, you should compute tenth roots, and nine powers of each. – Oct 03 '24 at 09:13
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is 1.1472 = 3^0.10? because if 3^a+b+c... = (3^a)(3^b)(3^c)...., so (3^3) ( 3^0.1) ( 3^0.04) = 3^3+0.1+0.04... which gives 3^π? – Jahraa Kabir Rose Oct 03 '24 at 11:23
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Um actually I am an O Level candidate 2026, CAIE, I have only learnt binary addition, two's complement, binary to decimal and decimal to binary conversions only in computer science subject. But I never heard about binary numeration. – Jahraa Kabir Rose Oct 04 '24 at 12:20
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If you could explain more easily it would be a great help. I also have later read Heron's method, it kinda cleared things up. – Jahraa Kabir Rose Oct 04 '24 at 12:23