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While messing around with Wolfram Alpha, I generated a table of the first 100 terms of $f\left(x\right)=\frac{x}{x+1}$ and applied the PowerExpand[] function. I noticed right away that among the 'unfriendly' values there was the occasional rational. I also noticed that these rationals were all of the form $\frac{x}{x+1}$. So, I looked for a pattern in the intervals and realized that $f(x)$ is rational when $x$ can be expressed as $2^n$, where $n$ is a natural number. Therefore, $f(2^x)$ yields the same values as $\frac{x}{x+1}$ over the integers. What I can't figure out is why any $b$-value other than the obvious $b=2$ would make $\frac{\log\left(2^{x}\right)}{\log\left(2^{x+1}\right)}$ equivalent to $\frac{x}{x+1}$. I hope the answer isn't obvious :)

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Due to the power property of logarithms where $\log_b(a^c) = c\log_b(a)$ (e.g., as shown in the third line, Power, of the table in the Product, quotient, power, and root section of Wikipedia's "Logarithm" article), you have

$$\begin{equation}\begin{aligned} \frac{\log_b(2^x)}{\log_b(2^{x+1})} & = \frac{x\log_b(2)}{(x+1)\log_b(2)} \\ & = \left(\frac{x}{x+1}\right)\left(\frac{\log_b(2)}{\log_b(2)}\right) \\ & = \frac{x}{x+1} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

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