# Why is $\frac{\log\left(2^{x}\right)}{\log\left(2^{x+1}\right)}$ equivalent to $\frac{x}{x+1}$ for any b-value?

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 :)

• Commented Mar 13, 2020 at 23:00

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{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}\tag{1}\label{eq1A}