# Why is it that $\frac{x^{\frac{1}{2^{15}}}-1}{0.000070271} \approx \log_{10}(x)$?

Recently, I came across this video, the method shown seemed good for a few logarithms. Then I tried to plot the equation $$\frac{x^{\frac{1}{2^{15}}}-1}{0.000070271}$$ and it looks exactly like the $$\log_{10}(x)$$ function. The result of both differs only by a very small fraction. How come it be such a great approximation of a logarithm, without the logarithm and how was this derived at all?

• I like the explanation in the video: "There's no particular scientific reason, it's just a method." Commented May 18, 2021 at 19:46
• I'm assuming you mean for relatively small x? As x goes to infinity, the polynomial will exceed the logarithm by an arbitrarily large percentage. Commented May 19, 2021 at 16:31
• Not that relatively small. Difference between $log(10^{100})$ and the given function with the same input is around $.035$
– Tim
Commented May 19, 2021 at 16:37
• $10^{100}$ is relatively small compared to infinity. Commented May 19, 2021 at 22:29

Try differentiating $$a^x$$ using the definition of the derivative. You should find that $$\frac{d}{dx}(a^x)=a^x\cdot\lim_{h\to0}\frac{a^h-1}{h} \, . \\$$ But we also have $$\frac{d}{dx}(a^x)=\frac{d}{dx}(e^{x\ln a})=a^x\ln a\, ,$$ and comparing the two yields $$\ln a=\lim_{h \to 0}\frac{a^h-1}{h} \, .$$ This means that for small $$h$$, $$\ln a \approx \frac{a^h-1}{h} \, .$$ The common logarithm is simply the natural logarithm scaled by a constant, and so we can write $$\log_{10}(x)=\frac{\ln x}{\ln 10}\approx\frac{x^h-1}{h\ln10}$$ For $$h=\dfrac{1}{2^{15}}$$, we obtain $$\log_{10}(x)\approx\frac{x^{2^{-15}}-1}{2^{-15}\cdot\ln10}=\frac{x^{2^{-15}}-1}{0.000070269\dots}$$ which is roughly the same as the approximation you have used in your question.

– Tim
Commented May 18, 2021 at 20:32
• @Tim: No problem. It was nice to see that the other answers gave different ways of obtaining the same result. Just to be clear: are you interested in why the method shown in the video works, or are you just looking for a way to derive this approximation? Commented May 18, 2021 at 20:46
• Both, but they're very connected. It works because the derivation was correct.
– Tim
Commented May 19, 2021 at 16:26
• @Tim: Good to hear. I encourage you to accept one of the answers here by clicking the green arrows. Commented May 19, 2021 at 22:01

This can be derived using the approximation $$e^x \approx 1+x$$ for small $$x$$. We have

\begin{align*} x^{2^{-15}} &= e^{2^{-15} \ln x } \\ &\approx 1 + 2^{-15} \ln x \end{align*}

and so $$\ln x \approx 2^{15} (x^{2^{-15}} - 1)$$

or $$\log_{10}(x) = \ln x / \ln {10} \approx \frac{1}{2^{-15}\ln(10)} \left(x^{2^{-15}} - 1\right)$$

where $$2^{-15}\ln(10) = .00007026932\ldots.$$

This relies on the assumption that $$2^{-15} \ln x$$ is small, therefore it will only work for $$x \ll e^{2^{15}}$$ although that is quite large of course.

• It does occur to me that both of our answers use the value of $\ln(10)$ to approximate the base-$10$ logarithm. I hope Tim will forgive us. Commented May 18, 2021 at 20:04
• @JoeLamond Ha, are you concerned with some circularity? Sure, but we just need a one-time lookup. :) Commented May 18, 2021 at 20:08
• Thank you for answering, I did think it did involve a bit of Taylor series. People here are so willing to help, wow. Great forum!
– Tim
Commented May 18, 2021 at 20:34
• @Tim You are welcome. For me it is just a fun distraction from work during the day. Commented May 18, 2021 at 20:35

hint

You might know that for small $$X,$$

$$e^X-1\sim X$$

but

$$x^{\frac{1}{2^{15}}}=e^{\frac{1}{2^{15}}\log(x)\ln(10)}$$

• Makes sense. Thanks!
– Tim
Commented May 18, 2021 at 20:32

This kind of approximation arises from the general method of using identities to improve a simpler approximation. Here I will make clear how to obtain it via this technique.

### Example 1

Use approximation $$\exp(x) ≈ 1+x$$ for $$x ≈ 0$$ and identity $$\exp(x·2^k) = \exp(x)^{2^k}$$ for any $$x∈ℝ$$ and $$k∈ℤ$$. From the latter, we get $$\exp(\ln(x)·2^k) = x^{2^k}$$ for any $$x > 0$$. From the former we get $$\exp(\ln(x)·2^k) ≈ 1+\ln(x)·2^k$$ for $$\ln(x)·2^k ≈ 0$$. Substituting $$k=-15$$ yields the result you asked for.

### Example 2

Use approximation $$\ln(1+x) ≈ x$$ for $$x ≈ 0$$ and identity $$\ln(x^{2^k}) = \ln(x)·2^k$$ for any $$x>0$$ and $$k∈ℤ$$. From the former we get $$\ln(x^{2^k}) ≈ x^{2^k}-1$$ for $$x^{2^k} ≈ 1$$. Note that as $$k→-∞$$ we do have $$x^{2^k} → x^0 = 1$$. Again, substituting $$k=-15$$ yields what you want.