Prove that the function f(x) = log_{x}(x+1) decreases on the interval (1,+infinity).
How should I approach this proof? Thoughts?
So $\displaystyle \frac{d}{dx}f(x)=\frac{x\log(x) - (x+1)\log(x+1)}{x(x+1)\log(x)\log(x)}$.
Since we need to show that $\displaystyle f'(x)<0,~\forall x\in(1,\infty)$, I guess the next step would be to consider the numerator and denominator separately. Consider the denominator first:
$\displaystyle x(x+1)\log(x)\log(x)$ is always positive since $\displaystyle x, x+1, \log(x)$ are by definition positive over the interval $\displaystyle (1,\infty)$.
It remains to show that the numerator is negative over the whole interval, i.e., $\displaystyle x\log(x) - (x+1)\log(x+1) < 0$. In other words we need to show that $\displaystyle x\log(x) < (x+1)\log(x+1)$.
I am not sure how to actually show this, since it is so obvious to me by looking at the graph. I thought about trying some kind of induction but we are not operating over N so it would be hard. Any further guidance you can give me?
You are on the right track. Use the rules you know about logarithms: $\displaystyle a\ln b = \ln b^a$ and then try $\displaystyle e^{\mbox{LHS}}$ compared to $\displaystyle e^{\mbox{RHS}}$ where LHS = Left Hand Side and RHS = Right Hand Side. How does it alter the inequality?
Here is my complete proof, does this work:
So $\displaystyle \frac{d}{dx}f(x)=\frac{x\log(x) - (x+1)\log(x+1)}{x(x+1)\log(x)\log(x)}$.
Since we need to show that $\displaystyle f'(x)<0,~\forall x\in(1,\infty)$, I guess the next step would be to consider the numerator and denominator separately. Consider the denominator first:
$\displaystyle x(x+1)\log(x)\log(x)$ is always positive since $\displaystyle x, x+1, \log(x)$ are by definition positive over the interval $\displaystyle (1,\infty)$.
It remains to show that the numerator is negative over the whole interval, i.e., $\displaystyle x\log(x) - (x+1)\log(x+1) < 0$. In other words we need to show that $\displaystyle x\log(x) < (x+1)\log(x+1)$.
$\displaystyle x\log(x) < (x+1)\log(x+1)$
$\displaystyle \log(x)^x < \log(x+1)^{x+1}$
$\displaystyle e^{\log(x)^x} < e^{\log(x+1)^{x+1}}$
Note: $\displaystyle e^{\log(x)} = x$, so we have:
$\displaystyle x^x < (x+1)^{x+1}$
Now $\displaystyle x^x < (x+1)^{x+1}$ is clearly true (or do I need to show something more) so it follows that the numerator of $\displaystyle f'(x)$ is negative $\displaystyle \forall x\in(1,\infty)$. Hence $\displaystyle f(x)$ is decreasing over the whole interval.
That is up to you. You can use the fact that for any $\displaystyle 1<x$, $\displaystyle x^x = 1\cdot x^x < x\cdot x^x = x^{x+1} < (x+1)^{x+1}$ since $\displaystyle x<x+1$.
Edit:
If you want to use what Plato wrote, you would want to show that $\displaystyle \ln(x)$ is a strictly increasing function on $\displaystyle (1,\infty)$. That is obvious since $\displaystyle (\ln(x))^\prime = \dfrac{1}{x} > 0$ for any positive $\displaystyle x$. Hence, since $\displaystyle x<x+1$, $\displaystyle \ln(x)<\ln(x+1)$, and the rest of what Plato wrote follows.