If x and y are positive integers satisfying 3^x - 3^y = 702 and \sqrt{x + y} = 3, what is … — GRE General Test

If $x$ and $y$ are positive integers satisfying $3^x - 3^y = 702$ and $\sqrt{x + y} = 3$ , what is the value of $x^2 - y^2$ ?

A
$9$
B
$18$
$27$ Cevap
D
$45$
E
$81$

Cevap

The correct value of

x^2 - y^2

is 27.

Squaring

\sqrt{x + y} = 3

gives

x + y = 9

. Factoring

3^x - 3^y = 702

gives

3^y(3^{x-y} - 1) = 702

. Since

702 = 3^3 \times 26

and

(3^{x-y} - 1)

is coprime to 3, we deduce

3^y = 3^3 \implies y = 3

, and

3^{x-3} - 1 = 26 \implies 3^{x-3} = 27 \implies x = 6

. Substituting

x = 6

and

y = 3

into

x^2 - y^2

gives

36 - 9 = 27

Adım Adım Çözüm

Eliminate the radical from the given linear equation.

Squaring both sides of

\sqrt{x + y} = 3

gives

x + y = 9

Squaring both sides removes the square root operator to establish a linear relationship between

x

and

y

Factor out the common exponential term

3^y

from

3^x - 3^y = 702

3^y(3^{x-y} - 1) = 702

Since

x

and

y

are positive integers and

702 > 0

, it must be true that

x > y

, allowing factoring by exponent rules

3^x = 3^y \cdot 3^{x-y}

Find the prime factorization of 702 and match the power of 3.

702 = 27 \times 26 = 3^3 \times 26

, so

3^y(3^{x-y} - 1) = 3^3 \times 26

The factor

(3^{x-y} - 1)

is not divisible by 3 because

3^{x-y}

is a multiple of 3 for

x > y

. Therefore, all powers of 3 in 702 must belong to

3^y

Solve for the values of

y

and

x

y = 3

and

3^{x-3} - 1 = 26 \implies 3^{x-3} = 27 = 3^3 \implies x - 3 = 3 \implies x = 6

Equating prime component bases yields

y = 3

and

x = 6

, which satisfies

x + y = 6 + 3 = 9

Calculate the target expression

x^2 - y^2

x^2 - y^2 = 6^2 - 3^2 = 36 - 9 = 27

Substituting

x = 6

and

y = 3

into

x^2 - y^2

yields 27 (or using

(x+y)(x-y) = 9 \times 3 = 27

Anahtar Kavram

Factoring exponential expressions using prime factorization, radical simplification, and difference of squares.

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