In the system of equations below, x and y are real numbers. 4^x \cdot 8^y = 128 9^x \cdot … — SAT

In the system of equations below, $x$ and $y$ are real numbers.

4^x \cdot 8^y = 128

9^x \cdot 3^y = 243

What is the value of $x$ ?

Answer: 2

Answer

The value of

x

2

To solve the system of equations, express all terms in each equation using common bases. In the first equation,

4^x \cdot 8^y = 128

can be written as

(2^2)^x \cdot (2^3)^y = 2^7

, which simplifies to

2^{2x+3y} = 2^7

, meaning

2x + 3y = 7

. In the second equation,

9^x \cdot 3^y = 243

can be written as

(3^2)^x \cdot 3^y = 3^5

, which simplifies to

3^{2x+y} = 3^5

, meaning

2x + y = 5

. Subtracting the two equations gives

(2x + 3y) - (2x + y) = 7 - 5 \implies 2y = 2 \implies y = 1

. Substituting

y = 1

back into

2x + y = 5

gives

2x + 1 = 5 \implies 2x = 4 \implies x = 2

Step-by-Step Solution

Rewrite the first equation using a common base of 2.

2x + 3y = 7

By converting

4^x

(2^2)^x = 2^{2x}

8^y

(2^3)^y = 2^{3y}

, and

128

2^7

, we can equate the exponents:

2x + 3y = 7

Rewrite the second equation using a common base of 3.

2x + y = 5

By converting

9^x

(3^2)^x = 3^{2x}

, keeping

3^y

, and converting

243

3^5

, we can equate the exponents:

2x + y = 5

Solve the system of linear equations for

y

y = 1

Subtracting

2x + y = 5

from

2x + 3y = 7

eliminates

x

, leaving

2y = 2

, which gives

y = 1

Substitute

y = 1

into one of the linear equations to solve for

x

x = 2

Substituting

y = 1

into

2x + y = 5

yields

2x + 1 = 5

, which simplifies to

2x = 4

, so

x = 2

Key Concept

Solving systems of exponential equations by converting to a common base and applying exponent laws.

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