TITLE: Why is it that $ \frac{x^{\frac{1}{2^{15}}}-1}{0.000070271} \approx \log_{10}(x)$?
QUESTION [11 upvotes]: 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?

REPLY [2 votes]: 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.