If `2^(x+y)=128` and `4^(x-y)=16`, then find `x/y`.

Options

A. `2/3`
B. `5/9`
C. `9/5`
D. `3/5`

Explanation

To solve the equations \(2^{(x+y)} = 128\) and \(4^{(x-y)} = 16\), we will follow these steps:

### Step 1: Rewrite the equations in terms of base 2
The first equation is \(2^{(x+y)} = 128\). We know that \(128\) can be expressed as a power of \(2\):
\[ 128 = 2^7 \]
Thus, we can rewrite the first equation as:
\[ 2^{(x+y)} = 2^7 \]
Since the bases are the same, we can equate the exponents:
\[ x + y = 7 \quad \text{(Equation 1)} \]

### Step 2: Rewrite the second equation in terms of base 2
The second equation is \(4^{(x-y)} = 16\). We can express \(4\) and \(16\) as powers of \(2\):
\[ 4 = 2^2 \quad \text{and} \quad 16 = 2^4 \]
Thus, we can rewrite the second equation as:
\[ (2^2)^{(x-y)} = 2^4 \]
This simplifies to:
\[ 2^{2(x-y)} = 2^4 \]
Again, since the bases are the same, we equate the exponents:
\[ 2(x - y) = 4 \]
Dividing both sides by 2 gives:
\[ x - y = 2 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of two equations:
1. \(x + y = 7\) (Equation 1)
2. \(x - y = 2\) (Equation 2)

We can solve these equations by adding them together:
\[ (x + y) + (x - y) = 7 + 2 \]
This simplifies to:
\[ 2x = 9 \]
Dividing by 2 gives:
\[ x = \frac{9}{2} \]

### Step 4: Substitute to find \(y\)
Now we can substitute \(x\) back into Equation 1 to find \(y\):
\[ \frac{9}{2} + y = 7 \]
Subtracting \(\frac{9}{2}\) from both sides:
\[ y = 7 - \frac{9}{2} \]
Converting \(7\) to a fraction:
\[ y = \frac{14}{2} - \frac{9}{2} = \frac{5}{2} \]

### Step 5: Find \(x/y\)
Now we can find the ratio \(x/y\):
\[ \frac{x}{y} = \frac{\frac{9}{2}}{\frac{5}{2}} = \frac{9}{2} \times \frac{2}{5} = \frac{9}{5} \]

Thus, the final answer is:
\[ \frac{x}{y} = \frac{9}{5} \]