Exponents and Algebraic Manipulation

Algebra Level 1

{ 2 a = x 3 a = y x 2 + y 2 + 4 x y = 72 + ( x + y ) 2 \large \begin{cases} 2^a = x \\ 3^a = y \\ x^2+y^2 + 4xy = 72 + (x+y)^2 \end{cases}

Given the above, find the value of a a .


The answer is 2.

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Kevin Lee by Kevin Lee , 31, USA

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1 solution

Chew-Seong Cheong
Jul 19, 2016

Given that { x = 2 a . . . ( 1 ) y = 3 a . . . ( 2 ) x 2 + y 2 + 4 x y = 72 + ( x + y ) 2 . . . ( 3 ) \begin{cases} x = 2^a & ...(1) \\ y = 3^a & ...(2) \\ x^2+y^2 + 4xy = 72 + (x+y)^2 & ...(3) \end{cases}

( 3 ) : x 2 + y 2 + 4 x y = 72 + ( x + y ) 2 x 2 + y 2 + 4 x y = 72 + x 2 + y 2 + 2 x y ⟹ 2 x y = 72 x y = 36 Substituting x = 2 a , y = 3 a ( 2 a ) ( 3 a ) = 36 6 a = 6 2 ⟹ a = 2 \begin{aligned} (3): \quad x^2+y^2 + 4xy & = 72 + (x+y)^2 \\ x^2+y^2 + 4xy & = 72 + x^2+y^2 + 2xy \\ \implies 2 xy & = 72 \\ \color{#3D99F6}{xy} & = 36 & \small \color{#3D99F6}{\text{Substituting }x=2^a, \ y=3^a} \\ (2^a)(3^a) & = 36 \\ 6^a & = 6^2 \\ \implies a & = \boxed{2} \end{aligned}

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hey I doubt this solution.. the answer is 1

Uchechukwu Okorie - 4 years, 10 months ago

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If a = 1 a=1 , then x = 2 x=2 and y = 3 y=3 , then L H S LHS = x 2 + y 2 + 4 x y = x^2+y^2+4xy = 4 + 9 + 24 = 37 = 4+9+24 = 37 but R H S RHS = 72 + ( x + y ) 2 = 72 + (x+y)^2 = 72 + 5 2 = 97 = 72+5^2 = 97 , ⟹ L H S ≠ R H S \implies LHS \ne RHS , therefore, a ≠ 1 a \ne 1 . You can try that with a = 2 a=2 .

Chew-Seong Cheong - 4 years, 10 months ago

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