Question

The real x and y satisfying simultaneously log8x + log4y2 = 5log8y + log4x2 = 7, then the value of xy equal to -(A) 29 (B) 212 (C) 218 (D) 224

Answer 1

is there any base given

Answer 2

Answer:

xy = +1

Step-by-step explanation:

Given that, log8x+log4y2=5

log8y+log4x2=7

So,log8x+log4y2=7

⇒log23x+log22y2=7

⇒13log2x+12log2y2=7

⇒log2(x)13+log2(y2)12=7

⇒log2[(x)13(y2)12]=7

⇒log2[(x)13y]=7

⇒(x)13y=27

⇒xy3=221      .....(1)And,

5log8y+log4x2=7

⇒5log23y+log22x2=7

⇒53log2y+12log2x2=7

⇒log2(y)53+log2(x2)12=7

⇒log2[(y)53(x2)12]=7

⇒log2[(y)53x]=7

⇒(y)53x=27

⇒y5x3=221      .....(2)Equating equation(2)and (1),

y2x2=1

⇒xy=±1