How many solutions $k>1$ does the equation $\exp ((k-1)/( k+1))=\sqrt{k}$ have?

Question

I have the following equation: $e^{\frac{k-1}{k+1}}=\sqrt{k}$. The question is: how many solutions does it have? ($e$ is Euler's constant and k is a positive real number >1).

Answer 1

Note that $$e^{(k-1)/(k+1)}=\sqrt k\iff \frac{k-1}{k+1}=\frac 12\log k\iff f(k)=-2$$ where $f(k)=(k+1)\log k-2k.$

Hence, the number of solutions in $e^{(k-1)/(k+1)}=\sqrt k$ is equal to that of $f(k)=-2.$

Since $f'(k)=\frac{g(k)}{k}$ where $g(k)=1-k+k\log k$, $g'(k)=\log k\gt 0$, we know that $g(k)$ is strictly increasing. With $g(1)=0$, we know $g(k)\gt 0$ for $k\gt 1$. So, we have $$f'(k)=\frac{g(k)}{k}\gt 0.$$

So, we know that $f(k)$ is strictly increasing for $k\gt 1$. With $f(1)=-2$, we know that there is no solution of $f(k)=-2.$

Answer 2

$\lim\limits_{k\rightarrow \infty} e^{\frac{k-1}{k+1}} = L\rightarrow\lim\limits_{k\rightarrow \infty} \frac{k-1}{k+1} = \ln L=1\rightarrow L=e$ yet $\sqrt{k}\rightarrow \infty$, both are monotonic increasing, therefore there is one root at x=1.

See this WolframAlpha output

Therefore, there are no roots >1

Answer 3

I think the only solution is at $k=1$ (outside your domain).

Take logarithms just to tidy up a bit, $$\frac{k-1}{k+1} = \frac{1}{2} \ln(k)$$ Note these agree at $k=1$. Compare derivatives of each quantity, $$\frac{1}{2k}$$ and $$\frac{2}{(k+1)^2}$$ Observe that these derivatives only agree at $k=1$ and afterwards the latter is smaller. Therefore they can never meet again.

Answer 4

Taking logarithms we compare $\frac{k-1}{k+1}=1-\frac{2}{k+1}$ and $\frac{1}{2}\ln(k)$ which are both increasing for $k > 0$, so they meet at most once. The value $\frac{1}{2}\ln(k)-(1-\frac{2}{k+1})$ at $k=1$ is $0$, so there is no solution for $k>1$ since the derivative is non-zero there.