- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,906

Recently I came across this problem "Solve the equation in $R$ of \(\displaystyle (x+1)^{x-1}= (x-1)^{x+1}\) and I only managed to solve it partially and here is what I have done:

\(\displaystyle (x+1)^{x-1}= (x-1)^{x+1}\)

\(\displaystyle \ln(x+1)^{x-1}=\ln(x-1)^{x+1}\)

\(\displaystyle (x-1)\ln(x+1)=(x+1)\ln(x-1)\)

\(\displaystyle x\ln(x+1)-\ln(x+1)=x\ln(x-1)+\ln(x-1)\)

\(\displaystyle x\ln \left(\frac{x+1}{x-1}\right)=\ln(x-1)(x+1)\)

And let $m=x-1$ we then have

\(\displaystyle (m+1)\ln \left(\frac{m+2}{m}\right)=\ln(m)(m+2)\)

\(\displaystyle m\ln \left(\frac{m+2}{m}\right)+\ln \frac{m+2}{m}=\ln(m)(m+2)\)

\(\displaystyle m\ln \left(\frac{m+2}{m}\right)=\ln \left((m)(m+2)\right)\left(\frac{m}{m+2}\right)\)

\(\displaystyle m\ln \left(\frac{m+2}{m}\right)=\ln m^2\)

\(\displaystyle \ln \left(\frac{m+2}{m}\right)^m=\ln m^2\)

By equating the argument from both sides, we obtain

\(\displaystyle \left(\frac{m+2}{m}\right)^m=m^2\)

The answer is then checked when $m=2$, i.e. $x=3$.

My question is, how do I determine that $-3$ is another answer to this problem and also, how do I make sure these are the only two answers to this problem?

Thanks in advance for any helps that anyone is going to pour in.