Task B-3.3. Solve the equation $x^{\log _{5} 6}-5 \cdot 6^{\log _{5} \sqrt{x}}=6$.
Solution. The equation makes sense for $x>0$.

Since $\log _{6} x=\frac{\log _{5} x}{\log _{5} 6}$ implies $\log _{6} x \cdot \log _{5} 6=\log _{5} x$, we have

$$
x^{\log _{5} 6}=\left(6^{\log _{6} x}\right)^{\log _{5} 6}=6^{\log _{6} x \cdot \log _{5} 6}=6^{\log _{5} x}
$$

Furthermore,

$$
6^{\log _{5} \sqrt{x}}=6^{\frac{1}{2} \log _{5} x}=\left(6^{\log _{5} x}\right)^{\frac{1}{2}}=\sqrt{6^{\log _{5} x}}
$$

Now we can rewrite the given equation in the form

$$
6^{\log _{5} x}-5 \cdot \sqrt{6^{\log _{5} x}}=6
$$

By substituting $\sqrt{6^{\log _{5} x}}=t$, we get the quadratic equation $t^{2}-5 t-6=0$ whose solutions are $t_{1}=6$ and $t_{2}=-1$.

Since $t>0$, we have only one solution.

$$
\sqrt{6^{\log _{5} x}}=6 \quad \Rightarrow \quad \log _{5} x=2 \quad \Rightarrow \quad x=25
$$