Problem 16. Solve the equation

$$
\lg \left(x^{2}+9\right)-3 \cdot 2^{x}+5=0
$$
Solution. The equation does not have negative roots because on $(-\infty ; 0)$ the function $P(x)=\lg \left(x^{2}+\right.$ $+9)+5$ is decreasing, the function $K(x)=3 \cdot 2^{x}$ is increasing, and $P(0)>K(0)$. To hypothesize about the number of roots of the given equation, we construct tables of the functions $P(x)$ and $K(x)$:

| $x$ | $P(x)$ | $K(x)$ | $x$ | $P(x)$ | $K(x)$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
|  |  |  |  |  |  |
| 0 | 5.954 | 3 | 1.6 | 6.062 | 9.09 |
| 0.2 | 5.956 | 3.45 | 1.8 | 6.087 | 10.4 |
| 0.4 | 5.961 | 3.95 | 2 | 6.113 | 12 |
| 0.6 | 5.971 | 4.55 | 2.2 | 6.141 | 13.8 |
| 0.8 | 5.984 | 5.22 | 2.4 | 6.169 | 15.8 |
| 1 | 6 | 6 | 2.6 | 6.197 | 18.2 |
| 1.2 | 6.018 | 6.89 | 2.8 | 6.226 | 20.9 |
| 1.4 | 6.039 | 7.92 | 3 | 5.255 | 24 |

Now we can hypothesize that if $x>2$, then $\lg \left(x^{2}+9\right)<2^{x}$ (this inequality is easily proven using the derivative). After this, it becomes clear that all roots of the given equation belong to the interval $[0 ; 2]$. The number 1 is a root of the equation.

From the table, it is clear that the intervals $[0 ; 0.8]$ and $[1.2 ; 2]$ cannot contain roots of the equation (1).

To prove that 1 is the only root of the equation (1), we construct tables of the functions $P^{\prime}(x)$ and $K^{\prime}(x)$:

| $x$ | $P^{\prime}(x)$ | $K^{\prime}(x)$ | $x$ | $P^{\prime}(x)$ | $K^{\prime}(x)$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0.8 | 0.072 | 3.6 | 1.1 | 0.093 | 4.4 |
| 0.9 | 0.079 | 3.8 | 1.2 | 0.099 | 4.7 |
| 1 | 0.080 | 4.1 |  |  |  |

Now it is clear that on the interval $(0.8 ; 1.2)$ the function $y=P(x)-K(x)$ is decreasing.