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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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    @Willie: I think J.P. meant that the limit for $x\to\infty$ is $1$.2011-04-04

1 Answers 1

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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}.

Now,

\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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    @Joriki Great suggestion. I edited the post accordingly.2011-04-04