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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

hint

You might know that for small $ X, $

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

but

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

Answer 4

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.