324. An interesting equation. Solve the equation

$$
x^{x+1}+x^{x}=1
$$
324. By taking the logarithm of both sides of the given equation, we obtain the equivalent equation

$$
f(x)=x \ln x+\ln (x+1)=0
$$

Since $\ln (1+x)>0$, then for $0<x<\frac{1}{e}$

$$
f^{\prime}(x)=\ln x+1+\frac{1}{1+x}=\ln x e+\frac{1}{1+x}>0
$$

and, therefore, $f(x)$ is monotonically increasing for $x>0$; hence, the given equation has no more than one real root. Approximate calculations yield the root value $x \approx 0.43605$.

$$
\text { [P. La Fratta, M. M., 44, } 56 \text { (January 1971).] }
$$