Решить уравнение. Ершова, С-25, вариант В2. №1(г) [latex] 20^{3x+2}= 4^{x+12} * 5^{5x-8} [/latex]

[latex]20^{3x + 2} = 4^{x + 12} \cdot 5^{5x - 8} \\ \\ 5^{3x + 2} \cdot 4^{3x + 2} = 4^{x + 12} \cdot 5^{5x - 8} \\ \\ \dfrac{5^{3x + 2} \cdot 4^{3x + 2}}{4^{x + 12} \cdot 5^{5x - 8}} = 1 \\ \\ 5^{3x + 2 - 5x + 8} \cdot 4^{3x + 2 - x - 12} = 1 \\ \\ 5^{-2x + 10} \cdot 4^{2x - 10} = 5^0 \cdot 4^0 \\ \\ -2x + 10 = 0 \ \ \ \ \ \ \ \ \ 2x - 10 = 0 \\ \\ x = 5 [/latex]

[latex]20^{3x+2}=4^{x+12}\cdot 5^{5x-8}\\\\5^{3x+2}\cdot 4^{3x+2}=4^{x+12}\cdot 5^{5x-8}\\\\5^{3x}\cdot 5^2\cdot 4^{3x}\cdot 4^2=4^{x}\cdot 4^{12}\cdot 5^{5x}\cdot 5^{-8}\; |:(4^{x}\cdot 5^{5x})\ne 0\\\\ \frac{5^{3x}\cdot 4^{3x}}{4^{x}\cdot 5^{5x}}\cdot 5^2\cdot 4^2=4^{12}\cdot 5^{-8}\\\\\frac{4^{2x}}{5^{2x}} = \frac{4^{12}}{5^8\cdot 5^2\cdot 4^2} \\\\(\frac{4}{5})^{2x}= \frac{4^{10}}{5^{10}} \\\\(\frac{4}{5})^{2x}=( \frac{4}{5} )^{10}\\\\2x=10\\\\x=5[/latex]