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Given $y=\ln(2x-1)/\ln(x)$, prove $y$ is decreasing for $x>1$.

While this is obvious by couple computations, the usual differentiation method to show this is true is not getting me anywhere since finding the $y'=0$ point is rather nuisance with ln and what not.

We know the limit of this function is 1, and with first few computations we can see that it does indeed decrease. However is this not an insufficient explanation? If there is a clean way of showing this, please do share!

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the limit $x\to 1$ of your function is not 1. (If it were 1 you'd have a problem: quite obviously $\ln (2x-1) > \ln x$ in the region you specified...) – Willie Wong Apr 4 '11 at 23:23
@Willie: I think J.P. meant that the limit for $x\to\infty$ is $1$. – joriki Apr 4 '11 at 23:32

The usual derivative method still works, despite having logs. Here, we find

$$ y' = \frac{\frac{2}{2x-1} \cdot \ln(x) - \ln(2x - 1) \cdot \frac{1}{x}}{(\ln x)^2}. $$


$$\begin{align}y' < 0 &\Longleftrightarrow \frac{2}{2x-1} \cdot \ln(x) - \ln(2x - 1) \cdot \frac{1}{x} < 0 \\ & \Longleftrightarrow \frac{2x \ln x}{x(2x - 1)} - \frac{(2x -1)\ln(2x-1)}{x(2x - 1)} < 0 \\ &\Longleftrightarrow 2x \ln x - (2x-1)\ln(2x-1) < 0 \end{align}.$$

Now, we just need to show that for all $x > 1$, we have

$$ F(x) := 2x \ln x - (2x-1)\ln(2x-1) < 0. $$

But, $F(1) = 0$, and so....

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I think it gets slightly easier if you put everything on a common (positive) denominator $x(2x-1)$ and just show that the numerator is negative for $x>1$. – joriki Apr 4 '11 at 23:19
@Joriki Great suggestion. I edited the post accordingly. – JavaMan Apr 4 '11 at 23:26

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