Romanian Mathematical Olympiad

The equation can be written ${\log }_2(1+\sqrt{x})=(2^x-1{)}^2$. The function $f:[0,\infty )\to [0,\infty )$, defined by $f(x)={\log }_2(1+\sqrt{x})$, is one to one and onto and its inverse $f^{-1}:[0,\infty )\to [0,\infty )$ is given by $f^{-1}(x)=(2^x-1{)}^2$. As $f$ is increasing strictly, the equation can be written $f(x)=f^{-1}(x)=x$, or equivalently $2^x=1+\sqrt{x}$. One can easily check that $x_1=0$ and $x_2=1$ are solutions. As $g:[0,\infty )\to [1,\infty )$, $g(x)=2^x$ is convex and the function $h:[0,\infty )\to [1,\infty )$, $h(x)=1+\sqrt{x}$ is concave, these are the only solutions.