Solve the following equations:
a) $64 x^{1.42}-7 x^{-0.16}=9 x^{3}$,
b) $3 x^{2}=13 x^{0.8}+10 x^{-0.4}$.
a) By rearranging and highlighting:

$$
x^{-0.16}\left(9 x^{3.16}-64 x^{1.58}+7\right)=0
$$

The highlighted factor cannot be 0, because the assumption $x^{-0.16}=1 / x^{0.16}=0$ leads to the contradiction $1=0$ with $x \neq 0$, and the negative exponent power is not defined for $x=0$. By substituting $x^{1.58}=y$ in the parentheses,

$$
9 y^{2}-64 y+7=0, \quad \text { and from this } \quad y_{1}=1 / 9, y_{2}=7
$$

Now, from the equation $x^{1.58}=1 / 9$, we get $\lg x=-(\lg 9): 1.58 \approx-0.6040=0.3960-1, x \approx 0.2489$; and from $x^{1.58}=7$, we get $x \approx 3.427$.

b) Similarly, from the second equation $x^{-0.4}\left(3 x^{2.4}-13 x^{1.2}-10\right)=0 ; x^{1.2}=z ; 3 z^{2}-13 z-10=0, z_{1}=-2 / 3, z_{2}=5$ From the equation $x^{1.2}=x^{\frac{6}{5}}=-\frac{2}{3}$, we do not get a solution, because $x=\sqrt[6]{\left(-\frac{2}{3}\right)^{5}}=\sqrt[6]{-\frac{32}{243}}$ is not a real number. From the equation $x^{1.2}=5$, we get $x \approx 3.824$.

Note. $x \approx-0.2489$ and $x \approx-3.427$, as well as $x \approx-3.824$, are not considered solutions, because the fractional exponent power is generally not defined for a negative base.