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CONTEXT:

I'm a Calculus I student and recently realized (While trying to solve an unrelated problem) that while I can add, substract, multiply or divide pretty much any two real numbers by hand to a pretty reasonable accuracy (< 0.05% error if I use three digits after the decimal).

I literally wouldn't know how to solve something like $(5.42)^{3.58}$ or $π^{π-1}$ to even 1 decimal digit (after the comma) of accuracy. I did learn about Taylor polynomials as approximations of functions recently, and have found them really useful for approximating (by hand), something like $\sin(0.64)$ or $\cos(0.13)$, but I haven't found a way to use them for exponential equations ($a^x$) without knowing the natural log of the number beforehand.

I searched online for how to solve $a^b$, but only found solutions for "clean numbers", or fractional exponents that end up simplifying the problem (via roots): e.g. $32^{0.4}$ = $32^\frac 25 = (32^{\frac 15})^2 = 2^2 = 4.$

Question: Is there a method to solving by hand something like $a^b$, where a and b are both real numbers, to a reasonable (Let's say < 0.1%) accuracy ?

insipidintegrator
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Anton Little
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  • $\ln a^b=b\ln a$ and there are good numerical methods for computing $\ln x$. And then there are good numerical methods for exponentiating $b\ln a$. – lulu Jun 26 '22 at 20:01
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    It would be very sensible to require $a$ to be positive - you do not want to be taking arbitrary powers of negative numbers. The traditional way of doing this "by hand" was use logarithm tables in base $10$ and reduce the question to addition with $(5.42)^{3.58} = 10^{3.58 \log_{10}(5.42)} = 10^{10^{\log_{10}(3.48)+\log_{10}(\log_{10}(5.42))}}$ – Henry Jun 26 '22 at 20:02
  • Specifically, using a precomputed "logarithm table". https://math.stackexchange.com/questions/1478490/how-did-john-napier-make-log-table https://math.stackexchange.com/questions/485028/how-was-the-first-log-table-put-together – DanielV Jun 26 '22 at 20:05
  • Many electrical engineers have memorized the common logs of integers for rough calculations of this sort. (We actually use decibels for this, e.g., multiplying by 2 adds 3 dB, dividing by 5 subtracts 7 dB) – Phil Freedenberg Jun 26 '22 at 20:20

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