We are tasked with solving the equation:

\[
2^x + 9^y = 59305
\]

To find the values of \(x\) and \(y\), and then compute \(x + y\).

### Step 1: Analyze the equation

We are given the sum of two exponential terms, \(2^x\) and \(9^y\), equating to 59305. Since 9 is a perfect square (\(9 = 3^2\)), we can rewrite the second term as:

\[
2^x + (3^2)^y = 59305
\]
\[
2^x + 3^{2y} = 59305
\]

### Step 2: Trial and error for reasonable values of \(x\) and \(y\)

Since exponential equations can be tricky to solve algebraically without more constraints, one approach is to test various integer values of \(x\) and \(y\) and check whether the equation holds.

Let’s calculate some powers of 2 and powers of 9 (or 3 raised to even powers), and check for values that could sum to 59305.

Let me calculate appropriate values.The values that satisfy the equation \( 2^x + 9^y = 59305 \) are \( x = 8 \) and \( y = 5 \).

### Step 3: Calculate \( x + y \)
\[
x + y = 8 + 5 = 13
\]

Thus, \( x + y = 13 \).

Would you like any additional details or explanations?

Here are some related questions:
1. What are some properties of exponential equations?
2. How can exponential equations be solved algebraically?
3. How does changing the base affect the growth rate of exponential functions?
4. Can this type of equation have multiple solutions?
5. How can logarithms be used to solve exponential equations?

**Tip:** When solving exponential equations by trial and error, start with reasonable small powers and check sums incrementally.

Discover how to solve the exponential equation 2^x + 9^y = 59305 using trial and error methods. This problem explores key concepts in exponential growth and is suitable for students in Grades 9-11. Detailed step-by-step calculations lead to the solution x + y = 13.

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