20^(3x+2)=4^(x+12)*5(5x-8) Помогите решить это показательная функция

20^(3x+2)= 4^(x+12)*5(5x-8) 4^(3x+2) * 5^(3x+2) =4^(x+12)*5(5x-8) 4²*4³ˣ * 5²*5³ˣ = 4¹²*4ˣ * 5⁻⁸*5⁵ˣ   сократим 4²ˣ*5¹⁰ =4¹⁰*5²ˣ 4²ˣ/5²ˣ = 4¹⁰/5¹⁰ (4/5)²ˣ=(4/5)¹⁰ 2х=10 х=5

[latex] 20^{3x+2} = 4^{x+12}5^{5x-8}, [/latex] [latex]4^{3x+2}5^{3x+2} = 4^{x+12}5^{5x-8}, [/latex] [latex]4^{3x}4^{2}5^{3x}5^{2} = 4^{x}4^{12}5^{5x}5^{-8}, [/latex] [latex]4^{2x}5^{10} = 4^{10}5^{2x},[/latex] [latex]( \frac{4}{5}) ^{2x} = ( \frac{4}{5}) ^{10}, [/latex] [latex]2x = 10,[/latex] [latex]x = 5.[/latex]