Solving the System of Equations with Exponents

We are given two equations involving variables x and y in the exponents. Our goal is to find the possible value(s) for the expression x + y.

Analyzing the First Equation

The first equation is given as:

\(4^x \times 2^y = 128\)

To solve this, we should express all terms with the same base. The base 2 is a common factor:

• We know that \(4 = 2^2\).
• We also know that \(128 = 2^7\) (since \(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 128\)).

Substituting these into the equation:

\((2^2)^x \times 2^y = 2^7\)

Using the exponent rule \((a^m)^n = a^{mn}\):

\(2^{2x} \times 2^y = 2^7\)

Using the exponent rule \(a^m \times a^n = a^{m+n}\):

\(2^{2x+y} = 2^7\)

Since the bases are equal, the exponents must be equal:

Equation A: \(2x + y = 7\)

Analyzing the Second Equation

The second equation is given as:

\(3^{3x} \times 3^{2y} - 9^{xy} = 0\)

First, rearrange the equation:

\(3^{3x} \times 3^{2y} = 9^{xy}\)

Now, express all terms with the same base. The base 3 is suitable:

• We know that \(9 = 3^2\).

Substituting this into the equation:

\(3^{3x} \times 3^{2y} = (3^2)^{xy}\)

Using the exponent rule \(a^m \times a^n = a^{m+n}\) on the left side and \((a^m)^n = a^{mn}\) on the right side:

\(3^{3x+2y} = 3^{2xy}\)

Since the bases are equal, the exponents must be equal:

Equation B: \(3x + 2y = 2xy\)

Solving the System of Equations

We now have a system of two equations:

1. \(2x + y = 7\) (Equation A)
2. \(3x + 2y = 2xy\) (Equation B)

From Equation A, we can easily express y in terms of x:

\(y = 7 - 2x\)

Substitute this expression for y into Equation B:

\(3x + 2(7 - 2x) = 2x(7 - 2x)\)

Expand and simplify both sides:

\(3x + 14 - 4x = 14x - 4x^2\)

\(-x + 14 = 14x - 4x^2\)

Rearrange the terms to form a quadratic equation:

\(4x^2 - x - 14x + 14 = 0\)

\(4x^2 - 15x + 14 = 0\)

Solving the Quadratic Equation for x

We can solve the quadratic equation \(4x^2 - 15x + 14 = 0\) by factoring. We look for two numbers that multiply to \(4 \times 14 = 56\) and add up to -15. These numbers are -7 and -8.

Rewrite the middle term:

\(4x^2 - 8x - 7x + 14 = 0\)

Factor by grouping:

\(4x(x - 2) - 7(x - 2) = 0\)

\((4x - 7)(x - 2) = 0\)

This gives us two possible values for x:

• Case 1: \(4x - 7 = 0 \implies 4x = 7 \implies x = \frac{7}{4}\)
• Case 2: \(x - 2 = 0 \implies x = 2\)

Finding the Corresponding Values for y and x + y

Now we find the value of y for each value of x using Equation A: \(y = 7 - 2x\).

Case 1: If \(x = 2\)

\(y = 7 - 2(2) = 7 - 4 = 3\)

Let's check if \((x, y) = (2, 3)\) satisfies Equation B: \(3x + 2y = 2xy\)

Left side: \(3(2) + 2(3) = 6 + 6 = 12\)

Right side: \(2(2)(3) = 12\)

The equation is satisfied. For this case, \(x + y = 2 + 3 = 5\).

Case 2: If \(x = \frac{7}{4}\)

\(y = 7 - 2(\frac{7}{4}) = 7 - \frac{14}{4} = 7 - \frac{7}{2} = \frac{14}{2} - \frac{7}{2} = \frac{7}{2}\)

Let's check if \((x, y) = (\frac{7}{4}, \frac{7}{2})\) satisfies Equation B: \(3x + 2y = 2xy\)

Left side: \(3(\frac{7}{4}) + 2(\frac{7}{2}) = \frac{21}{4} + 7 = \frac{21}{4} + \frac{28}{4} = \frac{49}{4}\)

Right side: \(2(\frac{7}{4})(\frac{7}{2}) = \frac{2 \times 7 \times 7}{4 \times 2} = \frac{98}{8} = \frac{49}{4}\)

The equation is satisfied. For this case, \(x + y = \frac{7}{4} + \frac{7}{2} = \frac{7}{4} + \frac{14}{4} = \frac{21}{4}\).

Comparing with Options

We found two possible values for x + y: 5 and \(\frac{21}{4}\) (which is 5.25).

Let's look at the given options:

• 7
• 5
• 3
• 1

The value 5 is present among the options.

Conclusion

Based on the analysis of the two equations, one possible value for \(x+y\) is 5.

Revision Table: Solving Exponential Equations

Additional Information: Exponent Rules and Solving Systems

Solving equations involving exponents often relies on converting terms to the same base. Once bases are the same on both sides of an equality, you can equate the exponents. Here are some key exponent rules used:

• Product Rule: \(a^m \times a^n = a^{m+n}\)
• Power Rule: \((a^m)^n = a^{mn}\)
• Quotient Rule: \(\frac{a^m}{a^n} = a^{m-n}\) (where \(a \ne 0\))
• Zero Exponent Rule: \(a^0 = 1\) (where \(a \ne 0\))
• Negative Exponent Rule: \(a^{-n} = \frac{1}{a^n}\) (where \(a \ne 0\))

In this problem, after simplifying the exponential equations, we ended up with a system containing a linear equation and a non-linear equation. A common method to solve such a system is substitution:

• Solve the linear equation for one variable (like y in terms of x).
• Substitute this expression into the non-linear equation.
• This reduces the non-linear equation to a single variable, often resulting in a quadratic equation, as seen in this case.
• Solve the single-variable equation.
• Substitute the found values back into the linear equation to find the corresponding values of the other variable.

Always check your solutions by substituting the pairs (x, y) back into the original equations to ensure they satisfy both.