Question

solve for X and Y : 2^x+1 = (√32)^y and 81^4-x/2 - 27^y = 0

Answer 1

Answer:

x=37/8, y=18/8=9/4

Step-by-step explanation:

$2^{x+1}=\sqrt{32}^y\\2^{x+1}=2^{\frac{5}{2} *y}\\log_2(2^{x+1})=log_2(2^\frac{5y}{2} )\\x+1=5y/2\\2x+2=5y......................(i)\\\\81^{4-\frac{x}{2} }-27^y=0\\(3^4)^{4-\frac{x}{2} }=(3^3)^y\\log_3(3^{16-2x})=log_3(3^{3y})\\16-2x=3y...........(ii)$

Adding (i) and (ii),

$2x+2+16-2x=5y+3y\\8y=18\\y=\frac{18}{8}\\ \\x=\frac{5y-2}{2} =\frac{5(\frac{18}{8} )-2}{8}\\ =\frac{37}{8}$

Answer 2

this is my answer hope it will help you

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