Помогите решить уравнение ЕГЭ: 1+log1/3(10-x)=log1/3(4-x)

Вместо ≤ или ≥ пиши просто: меньше или больше
$1 + log_{\frac{1}{3}}(10-x) = log_{\frac{1}{3}}(4-x) \\log^{\frac{1}{3}} _{\frac{1}{3}} + log_{\frac{1}{3}}(10 - x) = log_{\frac{1}{3}}(4-x) \\ ODZ: \left \{ {{ \left \{ {{(10-x)\geq 0} \atop {4-x \geq 0}} \right. } \atop {log_{\frac{1}{3}}( \frac{1}{3} (10 - x))= log_{\frac{1}{3}}(4-x) } }} \right. \\ \left \{ {{\left \{ {{x - 10 \leq 0} \atop {x - 4 \leq 0}} \right. } \atop { \frac{1}{3}(10 - x) = 4 - x}} \right.$
$\\ \left \{ {{ \left \{ {{x \leq 10} \atop {x \leq 4}} \right. } \atop {10 - x = 12 - 3 x}} \right. \\ \left \{ {{x \leq 4} \atop {2x=2}} \right. \left \{ {{x \leq 4} \atop {x=1}} \right. \\ OTVET: x = 1$