Solve the inequality $1+\sqrt{14-x} \geq \log _{2}(x-2)$ over the set of real numbers.
The function on the left side is only defined if $x \leq 14$, and the function on the right side is defined if $x>2$. Therefore, we must look for the solution in the interval $2<x \leq 14$.

Equality can hold, for example, if on the right side $(x-2)$ is some power of 2 (with an integer exponent), and on the left side, the expression under the square root, $14-x$, is a perfect square. The powers of 2 in the interval $(2 ; 14]$ are 4 and 8. If $x-2=4$, then $x=6$ and $14-6=8$ is not a perfect square.

If, however, $x-2=8$, then $x=10$ and $14-10=4$ is a perfect square, and this is indeed a solution to the equation.

Since the function on the left side is strictly decreasing in the interval $(2 ; 14]$, and $\log _{2}(x-2)$ is strictly increasing, equality cannot hold at any other point.

Let's plot the graphs of the functions.

The

graphs

show us

that the inequality holds exactly when $2<x \leq 10$.

Note. Many only drew the graphs of the functions and tried to read the result from them without any justification. These solutions are incomplete and are worth only 2 points. Similarly incomplete are those solutions that, apart from finding the point of equality $(x=10)$, did not mention the different monotonicity of the two sides.

![](https://cdn.mathpix.com/cropped/2024_05_02_14f1a273cd5f2d96d0cfg-1.jpg?height=312&width=656&top_left_y=782&top_left_x=729)