![image](https://tex.z-dn.net/?f=lg%28x%20%2B%201%29%20%2B%20lg%28x%20%2B%204%29%20%3D%201%20%5C%5C%3C%2Fp%3E%3Cp%3Elg%28%28x%20%2B%201%29%28x%20%2B%204%29%29%20%3D%201%20%5C%5C%3C%2Fp%3E%3Cp%3Elg%28x%5E2%20%2B%205x%20%2B%204%29%20%3D%201)
lg((x + 1)(x + 4)) = 1 \\
lg(x^2 + 5x + 4) = 1" alt="lg(x + 1) + lg(x + 4) = 1 \\
lg((x + 1)(x + 4)) = 1 \\
lg(x^2 + 5x + 4) = 1" align="absmiddle" class="latex-formula">
Потенциируем:
![image](https://tex.z-dn.net/?f=x%5E2%20%2B%205x%20%2B%204%20%3D%2010%5E1%20%3D%2010%20%5C%5C%3C%2Fp%3E%3Cp%3Ex%5E2%20%2B%205x%20-%206%20%3D%200%20%5C%5C%3C%2Fp%3E%3Cp%3E%28x%20%2B%206%29%28x%20-%201%29%20%3D%200%20%5C%5C%3C%2Fp%3E%3Cp%3Ex_1%20%3D%201%20%5C%5C%3C%2Fp%3E%3Cp%3Ex_2%20%3D%20-6)
x^2 + 5x - 6 = 0 \\
(x + 6)(x - 1) = 0 \\
x_1 = 1 \\
x_2 = -6" alt="x^2 + 5x + 4 = 10^1 = 10 \\
x^2 + 5x - 6 = 0 \\
(x + 6)(x - 1) = 0 \\
x_1 = 1 \\
x_2 = -6" align="absmiddle" class="latex-formula">
не подходит по ОДЗ, следовательно, ответ ![x = x_1 = -1 x = x_1 = -1](https://tex.z-dn.net/?f=x%20%3D%20x_1%20%3D%20-1)