Problem 18. Find the roots of the equation $F(x)=$ $=2^{2 x}-5 \cdot 3^{x}+29=0$ (to the nearest 0.01).
Solution. The given equation is equivalent to the equations:

$$
4^{x}+29=5 \cdot 3^{x}, \ln \left(4^{x}+29\right)=\ln 5+x \ln 3
$$

Let $P(x)=\ln \left(4^{x}+29\right)-x \ln 3-\ln 5$.

Solving the equation $P^{\prime}(x)=0$, we get $4^{x}=29 \ln 3 / \ln \frac{4}{3}$. Hence, $x \ln 4=\ln \frac{29 \ln 3}{\ln (4 / 3)}$ and $x_{0} \approx 3.395578$ (the root $x_{0}$ of the equation $P(x)=0$ is obtained using a calculator).

We find:

$$
F\left(x_{0}\right) \approx-68.73277 ; F(0)=29 ; F(6)=480
$$

Now it is clear that the equation $F(x)=0$ has two roots. Moreover, $x_{1}=2, x_{0}<x_{2}<6$. After refinement, we get $x_{2} \approx 5.55( \pm 0.01)$.

## Task 11. METHOD OF PARALLEL PROJECTIONS