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I was recently watching a video on Karatsuba's fast multiplication algorithm and the narrator stated something that intrigued me:

$2^{1.6} \approx 3 $

Specifically, I wondered what power $2$ would need to be raised to in order to be equal to $3$:

$2^p = 3$

Since my knowledge in the field of mathematics is rather lacking, I attempted to brute force the answer in the C# programming language:

double p = 1.5;
for (; Math.Pow(2, p) != 3;)
    p += 0.00000000000000000000000000000000000000000000000000000000000001;

Console.WriteLine(p);

Note: This snippet is intended for creating a CodeGolf.SE challenge, so please disregard the abuse of the for iterator here. It's abused for the sake of brevity.

However, as the incremental value increases in precision, the time it takes to reach an answer, also increases (likely by orders of magnitude), so this is a very inefficient way to determine the value needed for $p$.

What is the most effective way to determine the exponent needed in order for the following to be a true statement when $x$ and $z$ are known:

$$x^y = z$$

Lalit Tolani
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Hazel へいぜる
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    you want $\log_2(3)$ and we have the formula $\log_b(x) = \log(x)/\log(b)$ so you want $\log(3)/\log(2)$ – Asinomás Sep 01 '21 at 14:49
  • Use logarithms. There are good algorithms already out there for approximating logarithms to arbitrary precision. – Xander Henderson Sep 01 '21 at 14:50
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    "very inefficient way to determine the value needed for $p$" You can say that again... – JMoravitz Sep 01 '21 at 14:50
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    Certainly, a binary search would be faster than a linear search. But the real question is how do you calculate $2^x$ when $x$ is real? If you use $e^{x\log2},$ then you already know how to calculate $\log2.$ – Thomas Andrews Sep 01 '21 at 14:53
  • And don’t use $=3.$ You want to stop when $2^x>3.$ – Thomas Andrews Sep 01 '21 at 14:57
  • I will point out as well... thanks to floating point arithmetic, you might never get true equality between Math.pow(2,p) and 3 in your computer language... depending on precision and exactly where things lie... you might pass right over it and hit an infinite loop. Better to have used < rather than != here... – JMoravitz Sep 01 '21 at 14:58
  • Floating point arithmetic and exact equality are incredibly dangerous when programming in the first place. You might have for instance that 0.1+0.2 is not identically equal to 0.3 in a computer language. https://0.30000000000000004.com/ – JMoravitz Sep 01 '21 at 15:00

3 Answers3

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In general for positive reals $a$ and $b$ the value $x$ such that $b^x=a$ is known as $\log_b(a)$ and is called "the logarithm base $b$ of $a$" . In our case we have $b=2,a=3$.

The rule for converting logarithms with a special base to something with normal natural logarithm is $\log_b(a) = \log(a)/\log(b)$.

This is because we need $b^x=a$, and we have $b^x=(e^{\log(b)})^x = e^{\log(b)x}$ so we need $\log(b)x = \log(a)$.

In conclusion $\log_2(3) = \log(3)/\log(2) \approx 1.58496250072$

Asinomás
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If you don't want to use logarithms:

2^x = 3. x is obviously at least 1 but less than 2. So we create a sum starting with 1, and divide by 2^1: 2^y = 1.5. Square both sides, getting 2^(2y) = 2.25.

Obviously 1 <= 2y < 2. Add 0.5 to the sum, divide by 2, getting 2^(2z) = 1.125. Square to get 2^(4z) = 1.265625, square again to get 2^(8z) = 1.6018, square again to get 2^(16z) = 2.5658. So add 1/16 to the sum, divide by 2 getting 2^(16u) = 1.2829. And so on.

gnasher729
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One approach is to write $\log_2{3}$ as a continued fraction. You can do all these steps strictly in integer arithmetic, although quickly they become very large integers.

$$\begin{align} \log_{2}3&=1+\log_{2}(3/2)=1+\frac{1}{\log_{3/2}2}\\ \log_{3/2}2&=1+\log_{3/2}(4/3)=1+\frac1{\log_{4/3}(3/2)}\\ \log_{4/3}(3/2)&=1+\log_{4/3}(9/8)\\ \log_{9/8}(4/3)&=2+\log_{9/8}(2^8/3^5)\\ \log_{2^8/3^5}(9/8)&=2+\log_{2^8/3^5}(3^{12}/2^{19})\\&\vdots \end{align}$$

On the next step, you’d want to find the largest $n$ such that $$3^{12n+5}<2^{19n+8}$$ As you can see, just integer arithmetic, but we’re talking big integers. Indeed, the exponents grow exponentially.

This gives a sequence of approximations: $$1,2,\frac32,\frac85,\frac{19}{12},\frac{65}{41},\frac{84}{53},\frac{485}{306},\dots$$

Aside: The continued fractions for $\log_23$ are related to music. The pentatonic scale and the $12$-note scale sound good to our ears because $5$ and $12$ are denominators for good approximations for $\log_23,$ which means the scales can divide an octave into $5$ or $12$ roughly even intervals and still get almost perfect “fifths.”

Thomas Andrews
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