9.157. $\log _{0.5}(x+3)<\log _{0.25}(x+15)$.
## Solution.

Let's switch to base 2. We have $\log _{0.5}(x+3) 0 , } \\
{ x + 15 > 0 , } \\
{ x + 3 > \sqrt { x + 15 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x > -3, \\
x > -15, \\
x^{2} + 6x + 9 > x + 15
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x > - 3 , } \\
{ x ^ { 2 } + 5 x - 6 > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x > -3 \\
{\left[\begin{array}{l}
x > 1 \\
x < -6
\end{array}\right.}
\end{array}\right.\right.
\end{aligned}
$$

Using the number line, we have $x \in (1 ; \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0676.jpg?height=107&width=444&top_left_y=144&top_left_x=434)

Answer: $x \in (1 ; \infty)$.