Question

25^x=5^y-1 and [32]^y=4*8^x solve for X and y

Answer 1

$HEYA \: \\ \\ 25 {}^{x} = 5 {}^{(y - 1)} \\ and \\ (32) {}^{y} = 4 \times 8 {}^{x} \\ \\ 5 {}^{2x} = 5 {}^{(y - 1)} \\ compare \: powers \: of \: 5 \: we \: have \\ \\ 2x = y - 1 \\ \\ 2x + 1 = y \: \: ...Equation \: \: \: (i) \\ \\ (8 \times 4) {}^{y} = 4 \times 8 {}^{x} \\ \\ 8 {}^{(y - x)} = 4 {}^{(1 - y)} \\ \\ 2 {}^{3(y - x)} = 2 {}^{2(1 - y)} \\ \\ compare \: powers \: of \: 2 \: we \: have \: \\ \\ 3 y - 3x = 2 - 2y \\ \\ 5y - 3x = 2 \: ...Equation \: \: ...(ii) \\ \\ substitute \: value \: of \: y \: in \: equation \: ii \: we \: have \: \\ \\ 5(2x + 1) - 3x = 2 \\ \\ 7x = - 3 \\ \\ x = - 3 \div 7 \\ \\ and \: \\ \\ y = 1 \div 7$

Answer 2

Answer:

Step-by-step explanation:

   25^x = 5^y-1

=> 5^2x = 5^y-1

Therefore, 2x=y-1

             => x=(y-1)/2

   32^y= 4*8^x

=> 2^5y=2^(3x+2)

Therefore, 5y=3x+2=3(y-1)/2 + 2 (substitution)=(3y-3+4)/2=(3y+1)/2

              => 10y = 3y + 1

              => y=1/7

x=(y-1)/2 = -6/14 = -3/7