Solve the equation $4^{x}-3^{x}=\operatorname{tg} 15^{\circ}$.
Solution. To accurately calculate $\tan 15^{\circ}$, consider the right-angled triangle $ABC$ and the angle bisector of the $30^{\circ}$ angle at vertex $B$, which intersects the opposite leg at point $D$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4f3f70341c1ec6cd2ef8g-1.jpg?height=281&width=455&top_left_y=304&top_left_x=824)

The exact value of $\tan 15^{\circ}$ is the ratio of segments $CD$ and $BC$. For simplicity, let the hypotenuse of the right-angled triangle be 2 units, so the leg opposite the $30^{\circ}$ angle is 1 unit, and the adjacent leg is $\sqrt{3}$ units. Using the angle bisector theorem,

$$
CD = \frac{\sqrt{3}}{2 + \sqrt{3}} = \sqrt{3}(2 - \sqrt{3})
$$

Then we can immediately calculate that

$$
\tan 15^{\circ} = \frac{CD}{BC} = \frac{\sqrt{3}(2 - \sqrt{3})}{\sqrt{3}} = 2 - \sqrt{3}
$$

Thus, the exponential equation can be rewritten as:

$$
4^{x} - 3^{x} = 2 - \sqrt{3}, \quad 4^{x} - 3^{x} = 4^{\frac{1}{2}} - 3^{\frac{1}{2}}
$$

It is immediately clear that $x = \frac{1}{2}$ is a solution. We will show that this is the only solution. Since $3^{x}$ is positive for all exponents, we can divide both sides of the slightly rearranged equation by $3^{x}$:

$$
\left(\frac{4}{3}\right)^{x} = \frac{2 - \sqrt{3}}{3^{x}} + 1
$$

Consider the functions corresponding to the expressions on both sides of the equation. The function on the left side is strictly increasing because it is an exponential function with a base greater than 1. The function on the right side is the reciprocal of a strictly increasing, positive-valued function multiplied by a positive constant, so the function on the right side is strictly decreasing. Therefore, there can be at most one solution, which we have already found:

$$
x = \frac{1}{2}
$$