Need to solve the equation $$(x+1)^{x-1}=(x-1)^{x+1}$$ After applying logarithm on each side one obtains the following equation: $$f(x+1)=f(x-1)\text{, where }f(x)=\ln x/x $$ which doesn't seem to have a solution judging from the graph of $f$. What approach would you suggest? Thank you!
Asked
Active
Viewed 94 times
5
-
1draw a more careful graph by hand http://www.printablepaper.net/category/graph – Will Jagy Aug 26 '15 at 19:11
-
1Are complex numbers allowed, or are only real numbers allowed? – Scounged Aug 26 '15 at 19:11
-
4There is a solution at $x=3$. – J126 Aug 26 '15 at 19:12
-
1would like to get all the real solutions... – user264653 Aug 26 '15 at 19:20
1 Answers
2
Let's work with $$ (x-1)\ln(x+1)=(x+1)\ln(x-1) $$ and $x>1$. As my comment states, $x=3$ solves this. Consider $$ g(x)=(x-1)\ln(x+1)-(x+1)\ln(x-1). $$ Then $$ g'(x)=\ln\left(\frac{x+1}{x-1}\right)-\frac{4x}{x^2-1}. $$ Note that $\lim_{x\to \infty}g'(x)=0$. Also note that $$ g''(x)=\frac{2x^2+6}{(x^2-1)^2} $$ which is positive for $x>1$. Thus $g'(x)$ is an increasing function that approaches $0$. Thus $g'(x)<0$ for all $x>1$. Therefore $g(x)$ is decreasing for all $x>1$.
J126
- 17,451
-
1Sorry but I don't understand the last line: it says $g(x)<0$ for $x>1$ but $g(3)=0$. Thank you. – user264653 Aug 26 '15 at 19:47
-
1