If x and y are positive numbers with x > 1 that satisfy the system of equations below, wha… — SAT

If $x$ and $y$ are positive numbers with $x > 1$ that satisfy the system of equations below, what is the value of $x$ ?

x^y = 2^{12}

(2x)^{y-2} = 2^8

Answer: 8

Answer

The correct answer is 8.

The correct answer is 8. By manipulating the system of equations, we can express the second equation in terms of

x^y

. Substituting the value

2^{12}

for

x^y

allows us to express

x

in terms of

y

x = 2^{(y+2)/2}

. Substituting this back into the first equation results in the quadratic equation

y^2 + 2y - 24 = 0

. Since

y

must be positive, we find

y = 4

. Using this value of

y

, we solve for

x

and obtain

x = 8

, which is greater than 1, satisfying all given conditions.

Step-by-Step Solution

Distribute the exponent in the second equation:

(2x)^{y-2} = 2^8

2^{y-2} \cdot x^{y-2} = 2^8

Applying the power of a product rule:

(ab)^n = a^n b^n

Multiply both sides of the equation by

x^2

to align with the term

x^y

2^{y-2} \cdot x^y = 2^8 \cdot x^2

Since

x^{y-2} \cdot x^2 = x^{y-2+2} = x^y

, this allows us to substitute the value of

x^y

from the first equation.

Substitute

x^y = 2^{12}

into the equation.

2^{y-2} \cdot 2^{12} = 2^8 \cdot x^2

This uses the given system's first equation to eliminate the

x^y

term.

Simplify the left side and divide both sides by

2^8

to isolate

x^2

x^2 = 2^{y+2}

2^{y-2} \cdot 2^{12} = 2^{y+10}

. Dividing both sides by

2^8

yields

2^{y+10-8} = 2^{y+2}

Take the square root of both sides to find an expression for

x

x = 2^{\frac{y+2}{2}}

Since

x > 1

, we take the positive root:

x = (2^{y+2})^{1/2} = 2^{\frac{y+2}{2}}

Substitute the expression for

x

back into the first equation

x^y = 2^{12}

\left(2^{\frac{y+2}{2}}\right)^y = 2^{12}

This yields a single equation in terms of

y

Equate the exponents and solve the resulting quadratic equation for

y

y = 4

(since

y > 0

)

The equation is

2^{\frac{y^2+2y}{2}} = 2^{12} \implies \frac{y^2+2y}{2} = 12 \implies y^2+2y-24=0 \implies (y+6)(y-4)=0

. Since

y

must be positive,

y = 4

Substitute

y = 4

back into the expression for

x

x = 8

x = 2^{\frac{4+2}{2}} = 2^3 = 8

Key Concept

Solving systems of exponential equations by equating bases, applying exponent rules, and solving quadratic equations.

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