Question

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

Answer 1

$Big{32}^frac{x+5}{x-7}=0,25cdotBig{128}^frac{x+17}{x-3}\\Big(2^5Big)^frac{x+5}{x-7}=dfrac 14cdotBig(2^7Big)^frac{x+17}{x-3}\\Big2^{5cdotfrac{x+5}{x-7}}=2^{-2}cdotBig2^{7cdotfrac{x+17}{x-3}}\\Big2^{5cdotfrac{x+5}{x-7}}=Big2^{7cdotfrac{x+17}{x-3}-2}\\5cdotfrac{x+5}{x-7}=7cdotfrac{x+17}{x-3}-2~~~~bigg|cdot (x-7)(x-3)neq0\\5cdot(x+5)(x-3)=7cdot(x+17)(x-7)-2(x-7)(x-3)\\5cdot(x^2+2x-15)=7cdot(x^2+10x-119)-2(x^2-10x+21)\5x^2+10x-75=7x^2+70x-833-2x^2+20x-42\10x-75=70x-833+20x-42$

80x=800;    x = 10

Проверка :

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

Ответ : 10