11.8. Find the real values of $u$ and $v$ that satisfy the equation

$$
\left(u^{2020}-u^{2019}\right)+\left(v^{2020}-v^{2019}\right)=u \ln u+v \ln v .
$$
## Solution.

According to the domain of values of the equality in the statement, we must have $u>0$ and $v>0$. Fix $u>0$ and $v>0$; let the function $f: \square \rightarrow \square, f(x)=u^{x}+v^{x}$. This function is twice differentiable on $\square$. Calculating the first and second derivatives, we get $f^{\prime}(x)=u^{x} \ln u+v^{x} \ln v$ and $f^{\prime \prime}(x)=u^{x} \ln ^{2} u+v^{x} \ln ^{2} v$.

The equality in the statement becomes $f(2020)-f(2019)=f^{\prime}(1)$. Since the function $f$ is continuous on [2019,2020] and differentiable on $(2019,2020)$, by Lagrange's theorem, we obtain that there exists $c \in(2019,2020)$ such that $f^{\prime}(1)=f(2020)-f(2019)=f^{\prime}(c) \cdot(2020-2019)=f^{\prime}(c)$.

Furthermore, since the function $f^{\prime}$ is continuous on $[1, c]$ and differentiable on $(1, c)$ and $f^{\prime}(1)=f^{\prime}(c)$, by Rolle's theorem, we obtain that there exists $d \in(1, c)$ such that $f^{\prime \prime}(d)=0$.

We have obtained that $u^{d} \ln ^{2} u+v^{d} \ln ^{2} v=0$ for some $d \in(1, c)$. Since $u^{d} \ln ^{2} u \geq 0$ and $v^{d} \ln ^{2} v \geq 0$, it follows that $u^{d} \ln ^{2} u+v^{d} \ln ^{2} v=0$ only if $u^{d} \ln ^{2} u=v^{d} \ln ^{2} v=0$. Furthermore, since $u^{d}>0$ and $v^{d}>0$, we get $\ln u=\ln v=0$, i.e., $u=v=1$. Verifying these values, we obtain that $u=v=1$ are the only real values that satisfy the equality in the statement.