Question

32^(x+5/x-7) =0,25*128^(x+17/x-3)

Answer 1

$\Big{32}^\frac{x+5}{x-7}=0,25\cdot\Big{128}^\frac{x+17}{x-3}\\\\\Big(2^5\Big)^\frac{x+5}{x-7}=\dfrac 14\cdot\Big(2^7\Big)^\frac{x+17}{x-3}\\\\\Big2^{5\cdot\frac{x+5}{x-7}}=2^{-2}\cdot\Big2^{7\cdot\frac{x+17}{x-3}}\\\\\Big2^{5\cdot\frac{x+5}{x-7}}=\Big2^{7\cdot\frac{x+17}{x-3}-2}\\\\5\cdot\frac{x+5}{x-7}=7\cdot\frac{x+17}{x-3}-2~~~~\bigg|\cdot (x-7)(x-3)\neq0\\\\5\cdot(x+5)(x-3)=7\cdot(x+17)(x-7)-2(x-7)(x-3)\\\\5\cdot(x^2+2x-15)=7\cdot(x^2+10x-119)-2(x^2-10x+21)\\5x^2+10x-75=7x^2+70x-833-2x^2+20x-42\\10x-75=70x-833+20x-42$

80x=800;    x = 10

Проверка :

$\Big{32}^\frac{10+5}{10-7}=0,25\cdot\Big{128}^\frac{10+17}{10-3}\\\\\Big{32}^\frac{15}3=0,25\cdot\Big{128}^\frac{27}7\\\\32^5=\dfrac 14\cdot\Big(2^7\Big)^\frac{27}7\\\\\Big(2^5\Big)^5=2^{-2}\cdot 2^{27}\\\\2^{25}=2^{25}$

Ответ : 10