Sum of values of `x` and `y` satisfying `3^x-4^y=77, 3^(x/2)-2^y=7` is

Answer

To solve the equations \(3^x - 4^y = 77\) and \(3^{(x/2)} - 2^y = 7\), we can follow these steps:

### Step 1: Rewrite the equations
We start with the two equations:
1. \(3^x - 4^y = 77\)
2. \(3^{(x/2)} - 2^y = 7\)

### Step 2: Express \(4^y\) in terms of \(3^x\)
From the first equation, we can express \(4^y\):
\[ 4^y = 3^x - 77 \]

### Step 3: Express \(3^{(x/2)}\) in terms of \(2^y\)
From the second equation, we can express \(3^{(x/2)}\):
\[ 3^{(x/2)} = 2^y + 7 \]

### Step 4: Substitute \(3^{(x/2)}\) into the first equation
We know that \(3^x = (3^{(x/2)})^2\). Substituting \(3^{(x/2)}\) from Step 3 into this gives:
\[ 3^x = (2^y + 7)^2 \]

### Step 5: Expand the square
Expanding the square:
\[ 3^x = (2^y + 7)^2 = 2^{2y} + 14 \cdot 2^y + 49 \]

### Step 6: Set up the equation
Now, substitute \(4^y = 2^{2y}\) into the first equation:
\[ (2^{2y} + 14 \cdot 2^y + 49) - 2^{2y} = 77 \]
This simplifies to:
\[ 14 \cdot 2^y + 49 = 77 \]

### Step 7: Solve for \(2^y\)
Subtracting 49 from both sides:
\[ 14 \cdot 2^y = 28 \]
Dividing both sides by 14:
\[ 2^y = 2 \]

### Step 8: Solve for \(y\)
Taking the logarithm base 2:
\[ y = 1 \]

### Step 9: Substitute \(y\) back to find \(x\)
Now substitute \(y = 1\) back into the equation for \(3^{(x/2)}\):
\[ 3^{(x/2)} = 2^1 + 7 = 2 + 7 = 9 \]
Thus:
\[ 3^{(x/2)} = 3^2 \]
This implies:
\[ \frac{x}{2} = 2 \implies x = 4 \]

### Step 10: Find the sum \(x + y\)
Now we have \(x = 4\) and \(y = 1\). Therefore, the sum is:
\[ x + y = 4 + 1 = 5 \]

### Final Answer
The sum of values of \(x\) and \(y\) satisfying the equations is \(5\).

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