2. (10 points) When modeling the development of a certain process over time $t$, a physicist derived the dependence $y_{1}(t)=2^{-4 t}$, while a mechanic obtained the dependence $y_{2}(t)$ for the same process, which turned out to be the inverse function of $y_{1}(t)$. A mathematician ended the scientific debate by proving that both of these dependencies are sufficiently close to a hyperbolic function $y_{3}(t)=\frac{c}{t}$, where $c$ is some constant, and the equations $y_{1}=y_{3}$ and $y_{2}=y_{3}$ have exactly two solutions in this time interval, which can be easily guessed if $c$ is known.
1) Find this value of $c$ and the minimal interval $\left[t_{1}, t_{2}\right]$ where the mathematician's statement is valid.
2) Determine how many exact solutions the equation $y_{1}=y_{2}$ has.
## Solution.

$y_{1}(t)=2^{-4 t}=\left(\left(\frac{1}{2}\right)^{4}\right)^{t}=\left(\frac{1}{16}\right)^{t}$

$y_{2}(t)=\log _{\frac{1}{16}} t$

$y_{1}=y_{2} \Leftrightarrow\left(\frac{1}{16}\right)^{t}=\log _{\frac{1}{16}} t$

Since equation (1) involves inverse functions, it must have a solution on the bisector of the first quadrant $t_{*}=y_{*}$.

However, it is easy to verify (a guess that is obvious), that equation (1) has two more solutions:

For $t_{1}=\frac{1}{4}$ since $\sqrt[4]{\frac{1}{16}}=\log _{\frac{1}{16}}\left(\frac{1}{4}\right)=\frac{1}{2}$.

For $t_{2}=\frac{1}{2}$ since $\sqrt{\frac{1}{16}}=\log _{\frac{1}{16}}\left(\frac{1}{2}\right)=\frac{1}{4}$.

The constant $c$ in $y_{3}(t)$ is recovered for these symmetric solutions $\left(\frac{1}{4}, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, \frac{1}{4}\right)$ $y_{3}(t) \cdot t=\frac{1}{8}=c$

From the guessed solutions, it is clear that $\min \left[t_{1}, t_{2}\right]=\left[\frac{1}{4}, \frac{1}{2}\right]$

Answer: 1) $c=\frac{1}{8} ;\left[t_{1}, t_{2}\right]=\left[\frac{1}{4}, \frac{1}{2}\right] ; 2$ 2) 3 solutions (proven $t_{*}$ and guessed $t_{1}=\frac{1}{4}, t_{2}=\frac{1}{2}$ ).