We are given the equation:

\[
(2^a)^2 + 3^b = 13
\]

Let's simplify the expression step by step.

1. First, simplify \((2^a)^2\) as:

\[
(2^a)^2 = 2^{2a}
\]

So the equation becomes:

\[
2^{2a} + 3^b = 13
\]

Now, we need to find values of \(a\) and \(b\) that satisfy this equation. Since 13 is a relatively small number, we can try some integer values for \(a\) and \(b\).

### Testing possible values for \(a\) and \(b\):

- For \(a = 1\):

\[
2^{2(1)} + 3^b = 2^2 + 3^b = 4 + 3^b = 13
\]

This gives:

\[
3^b = 9
\]

Thus, \(b = 2\).

### Conclusion:
The values of \(a = 1\) and \(b = 2\) satisfy the equation.

The solution is:
\[
a = 1, \quad b = 2
\]

Would you like further details or explanations on this process? 

Here are 5 related questions for practice:
1. Solve the equation \((2^a)^2 + 5^b = 53\).
2. If \(2^{2a} + 3^b = 19\), what are the values of \(a\) and \(b\)?
3. Solve \(2^{a} + 3^{b} = 31\).
4. Find \(a\) and \(b\) for \(2^{2a} + 7^b = 50\).
5. What is the solution to \((3^a)^2 + 2^b = 28\)?

**Tip:** When solving equations involving exponents, it's often useful to try small integer values to test for possible solutions.

Explore the step-by-step process of solving the exponential equation (2^a)^2 + 3^b = 13, where the values of a and b are determined using algebraic methods and the laws of exponents. Suitable for students in Grades 9-10.

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