The real x and y satisfying simultaneously log8x + log4y2 = 5log8y + log4x2 = 7, then the value of - Maths - Relations and Functions

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

Priyanka Kedia · Accepted Answer

Note: Recheck your optionsGiven that,
log8x+log4y2=5log8y+log4x2=7So,log8x+log4y2=7⇒log23x+log22y2=7⇒13log2x+12log2y2=7⇒log2x13+log2y212=7⇒log2x13y212=7⇒log2x13y=7⇒x13y=27⇒xy3=221
1And,5log8y+log4x2=7⇒5log23y+log22x2=7⇒53log2y+12log2x2=7⇒log2y53+log2x212=7⇒log2y53x212=7⇒log2y53x=7⇒y53x=27⇒y5x3=221
2Equating equation2and 1,y2x2=1⇒xy=±1

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

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