If `3^(x-y) =27` and `3^(x+y) = 243`, then what is the value of x?

Answer

To solve the equations \(3^{(x-y)} = 27\) and \(3^{(x+y)} = 243\) for the value of \(x\), we can follow these steps:

### Step 1: Rewrite the equations in terms of powers of 3.
We know that:
- \(27 = 3^3\)
- \(243 = 3^5\)

So we can rewrite the equations as:
1. \(3^{(x-y)} = 3^3\)
2. \(3^{(x+y)} = 3^5\)

### Step 2: Set the exponents equal to each other.
Since the bases are the same, we can set the exponents equal to each other:
1. \(x - y = 3\) (from the first equation)
2. \(x + y = 5\) (from the second equation)

### Step 3: Solve the system of equations.
Now we have a system of two equations:
1. \(x - y = 3\)
2. \(x + y = 5\)

We can solve for \(x\) and \(y\) by adding the two equations together:
\[ (x - y) + (x + y) = 3 + 5 \]
This simplifies to:
\[ 2x = 8 \]
Now, divide both sides by 2:
\[ x = 4 \]

### Step 4: Find the value of \(y\).
Now that we have \(x\), we can substitute it back into one of the equations to find \(y\). Let's use \(x + y = 5\):
\[ 4 + y = 5 \]
Subtracting 4 from both sides gives:
\[ y = 1 \]

### Final Answer:
Thus, the value of \(x\) is \(4\).

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