Subject (4). Solve the equation $2^{x}+1=\left(3^{x}-1\right)^{\log _{2} 3}$.

EUGEN RADU[^2]
Solution. The domain $x \geq 0$ is forced by the fact that the exponent $\log _{2} 3$ is irrational, so the power of this exponent is not (well) defined for negative bases. Since $x=0$ is not a solution, the equation immediately transforms into the equivalent form $\log _{3}\left(2^{x}+1\right)=\log _{2}\left(3^{x}-1\right)$. For the common value $v$, we can write $2^{x}+1=3^{v}$ and $3^{x}-1=2^{v}$, from which $2^{x}+3^{x}=2^{v}+3^{v}$. But the function $t \mapsto 2^{t}+3^{t}$ is the sum of two increasing functions, hence increasing, which forces $v=x$. Then $2^{x}+1=3^{x}$, or $(2 / 3)^{x}+(1 / 3)^{x}=1$, with the left-hand side a decreasing function, thus having at most one real solution. Since $x=1$ satisfies, it is the unique solution.

Sure, the equation itself assumes that we are working with real numbers, but at grade IX we didn't bother to specify this in Subject 2. A minimum of coherence should be maintained.