Let's solve the system of equations:

\[
2^x + 2^y = 12
\]
\[
2^{x+y} = 32
\]

### Step 1: Solve the second equation
The second equation can be rewritten as:
\[
2^{x+y} = 2^5
\]
Since the bases are the same, we equate the exponents:
\[
x + y = 5
\]

### Step 2: Substitute \( y = 5 - x \) into the first equation
Substituting \( y = 5 - x \) into the first equation \( 2^x + 2^y = 12 \):
\[
2^x + 2^{5-x} = 12
\]

### Step 3: Solve the equation
To solve this equation, we test possible values of \( x \) that satisfy the equation.

- **For \( x = 2 \):**
  \[
  2^2 + 2^{5-2} = 4 + 8 = 12
  \]
  This works, so \( x = 2 \).

Substituting \( x = 2 \) into \( y = 5 - x \):
\[
y = 5 - 2 = 3
\]

Thus, the solution is \( x = 2 \) and \( y = 3 \).

### Final Answer:
\[
(x, y) = (2, 3)
\]

Would you like more details on any step, or do you have any questions?

Here are some related questions:
1. What if \( 2^{x+y} = 64 \)? How does this change the solution?
2. How can we generalize this method to other similar exponential systems?
3. What is the significance of solving the system using substitution versus elimination?
4. What happens if \( 2^x + 2^y = 16 \) and \( 2^{x+y} = 32 \)?
5. How would the solution change if the first equation were \( 2^x + 2^y = 14 \)?

**Tip:** When dealing with exponential equations, always consider equating the exponents after rewriting them in a comparable base format.

Explore how to solve the system of exponential equations 2^x + 2^y = 12 and 2^(x+y) = 32 step-by-step. Understand the manipulation of exponential equations and find the values of x and y that satisfy both equations. Suitable for Grades 9-12 students studying advanced algebra.

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