https://file.aiquickdraw.com/mathbot/file/17264336173323w2cpbid.jpg

The equation in the image is:

\[
3^{1 - x} = 27^{\frac{x + 1}{2}}
\]

We can solve this step by step.

### Step 1: Express 27 as a power of 3
Since \( 27 = 3^3 \), we can rewrite the equation as:

\[
3^{1 - x} = \left(3^3\right)^{\frac{x + 1}{2}}
\]

### Step 2: Apply the power of a power rule
The rule \((a^m)^n = a^{m \cdot n}\) allows us to simplify the right-hand side:

\[
3^{1 - x} = 3^{3 \cdot \frac{x + 1}{2}} = 3^{\frac{3(x + 1)}{2}}
\]

### Step 3: Set the exponents equal
Since the bases on both sides of the equation are the same (both are powers of 3), we can set the exponents equal:

\[
1 - x = \frac{3(x + 1)}{2}
\]

### Step 4: Solve for \( x \)
Now, we solve the equation \( 1 - x = \frac{3(x + 1)}{2} \).

1. Multiply both sides by 2 to eliminate the fraction:

\[
2(1 - x) = 3(x + 1)
\]

This simplifies to:

\[
2 - 2x = 3x + 3
\]

2. Move all the terms involving \( x \) to one side:

\[
-2x - 3x = 3 - 2
\]

This simplifies to:

\[
-5x = 1
\]

3. Solve for \( x \):

\[
x = -\frac{1}{5}
\]

### Final Answer:
The solution is \( x = -\frac{1}{5} \).

Would you like more details on any of these steps or have any questions?

Here are five related questions for further exploration:

1. How would you solve the equation if the base were different, such as 2 instead of 3?
2. What if the exponents were not directly equatable, such as having different bases like 2 and 5?
3. How can you check if your solution is correct by substituting it back into the original equation?
4. How do logarithms help when the bases are not the same?
5. What are some real-world applications of solving exponential equations?

**Tip:** When solving exponential equations with the same base, equate the exponents directly to simplify the process.

Solve the exponential equation 3^(1 - x) = 27^((x + 1)/2)

Learn how to solve the exponential equation 3^(1 - x) = 27^((x + 1)/2) using exponent rules and the power of a power rule. This solution involves setting the exponents equal and solving for x. Ideal for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation