Why does $\log_{10}(10^x + 1) - x$ approximate $\log_{10} e$?

Question

I was calculating base-$10$ logarithms and noticed that $\log_{10}(10^k+1) - k$ approaches $\log_{10} e$.

In the screenshot below, Log[a,b] calculates $\log_ab$, and N[p,q] gives the numerical value of $p$ to $q$ digits.

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Question 1: Why does $\log_{10}(10^x + 1)- x$ approximate $\log_{10} e$?

Question 2: Why $10^k+1$? I can't recognize the pattern when I try to calculate, e.g. $\log_{10}(10^k+2)$.

Answer 1

$$e = \lim_{n\to\infty}\left(1+\frac1n\right)^n$$

Replace $n$ with $10^x$, apply $\log_{10}$ to both sides, use continuity of logarithms, and then simplify using basic properties of logs.

As for your second question, putting a $2$ in the one's digit would then use $$e^2 = \lim_{n\to\infty}\left(1+\frac2n\right)^n$$ and then when you take logs, you get $\log_{10}(e^2)=2\log_{10}(e)$

Answer 2

As per the picture, you compute $$ \log_{10}(1+10^x)-x=\log_{10}(1+10^{-x})=\log_{10}e\cdot\ln(1+10^{-x}) $$ and $\ln(1+10^{-x})=10^{-x}-\frac1210^{-2x}+\frac1310^{-3x}+...$, which is exactly what you observe.

Answer 3

Since $\log_{10}(x)=\frac{\log(x)}{\log(10)}$ and $\log(1+x)=x+O\!\left(x^2\right)$, $$ \begin{align} \log_{10}\left(10^n+k\right)-n &=\log_{10}\left(1+k\,10^{-n}\right)\\[3pt] &=\frac{\log\left(1+k\,10^{-n}\right)}{\log(10)}\\ &=\frac{k\,10^{-n}+O\!\left(10^{-2n}\right)}{\log(10)}\\[6pt] &=k\log_{10}(e)\,10^{-n}+O\!\left(10^{-2n}\right) \end{align} $$

For example $$ \begin{align} \log_{10}\left(10^{\color{#C00}{50}}+\color{#090}{3}\right)-\color{#C00}{50}&=1.3028834457098\times10^{-\color{#C00}{50}}\\ \color{#090}{3}\,\log_{10}(e)&=1.3028834457098 \end{align} $$