5. The function $f(t)$ with domain $D(f)=[1,+\infty)$ satisfies the equation $f\left(\frac{t^{y}-4^{-3}}{2}\right)=y$ for any $y \geq 0$. For each value of $a \neq 0$, find all solutions to the inequality $f\left(\frac{a}{x-2 a}\right) \leq 1$. Solution: The function $g(y)=\frac{4^{y}+4^{-y}}{2}$ is monotonically increasing for $y \geq 0$, taking values in $[1 ;+\infty)$. By the condition, the function $f$ is the inverse of $g$, and thus is also monotonically increasing. Then

$$
f\left(\frac{a}{x+2 a}\right) \leq 1 \Longleftrightarrow \frac{a}{x+2 a} \leq g(1)
$$

under the condition that the function $f$ is defined, i.e., under the condition $\frac{a}{x+2 a} \geq 1$. It remains to solve the system

$$
\left\{\begin{array} { l } 
{ \frac { a } { x + 2 a } \leq \frac { 1 7 } { 8 } } \\
{ \frac { a } { x + 2 a } \geq 1 }
\end{array} \quad \Longleftrightarrow \left\{\begin{array}{l}
\frac{17 x+26 a}{x+2 a} \geq 0 \\
\frac{x+a}{x+2 a} \leq 0
\end{array}\right.\right.
$$

For $a>0$, we get

$$
\left\{\begin{array}{l}
x \in(-\infty ;-2 a) \cup\left[-\frac{26 a}{17} ;+\infty\right), \Longleftrightarrow x \in\left[-\frac{26 a}{17} ;-a\right] \\
x \in(-2 a ;-a]
\end{array}\right.
$$

For $a<0$, we get

$$
\left\{\begin{array}{l}
x \in\left(-\infty ;-\frac{26 a}{17}\right] \cup(-2 a ;+\infty), \\
x \in[-a ;-2 a)
\end{array} \Longleftrightarrow x \in\left[-a ;-\frac{26 a}{17}\right]\right.
$$
Answer: $\quad$ For $a>0: x \in\left[-\frac{26 a}{17} ;-a\right]$. For $a<0: x \in\left[-a ;-\frac{26 a}{17}\right]$.