7.249. $3 \cdot 4^{(x-2)}+27=a+a \cdot 4^{(x-2)}$. For what values of $a$ does the equation have a solution?
## Solution.

Rewrite the equation as

$$
\begin{aligned}
& 3 \cdot 4^{(x-2)}-a \cdot 4^{(x-2)}=a-27 \Leftrightarrow (3-a) \cdot 4^{(x-2)}=a-27 \Rightarrow \\
& \Rightarrow 4^{(x-2)}=\frac{a-27}{3-a}, \text{ where } \frac{a-27}{3-a}>0 .
\end{aligned}
$$

Taking the logarithm of both sides of this equation with base 4, we get $\log _{4} 4^{(x-2)}=\log _{4} \frac{a-27}{3-a} \Leftrightarrow x-2=\log _{4} \frac{a-27}{3-a}, x=2+\log _{4} \frac{a-27}{3-a}$, where $\frac{a-27}{3-a}>0$. Solving the resulting inequality using the interval method, we have

$$
\xrightarrow[+]{+} \stackrel{+}{+\sqrt{27}} a
$$

Thus, $a \in(3 ; 27)$.

Answer: $\quad 2+\log _{4} \frac{a-27}{3-a}$, where $a \in(3 ; 27)$.