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Part 1
Solve the exponential equation. Express irrational solutions in exact form.
left parenthesis two ninths right parenthesis Superscript x Baseline equals 8 Superscript 1 minus x

We are asked to solve the exponential equation:

\[
\left( \frac{2}{9} \right)^x = 8^{1 - x}
\]

### Step 1: Rewrite both sides in terms of powers of common bases

We notice that \( \frac{2}{9} \) and 8 can be expressed as powers of smaller numbers:
- \( \frac{2}{9} = \frac{2^1}{3^2} \), so we can express it as \( 2^1 \times 3^{-2} \).
- \( 8 = 2^3 \), so we can express \( 8^{1 - x} \) as \( (2^3)^{1 - x} = 2^{3(1 - x)} \).

Now the equation becomes:
\[
(2^1 \times 3^{-2})^x = 2^{3(1 - x)}
\]

### Step 2: Apply exponent rules

Use the property that \( (a \times b)^x = a^x \times b^x \) to separate the terms:
\[
2^x \times 3^{-2x} = 2^{3(1 - x)}
\]

### Step 3: Set up two equations by comparing powers of the same base

#### For the powers of 2:
Since we have \( 2^x = 2^{3(1 - x)} \), we can set the exponents equal to each other:
\[
x = 3(1 - x)
\]

Solve for \( x \):
\[
x = 3 - 3x
\]
\[
x + 3x = 3
\]
\[
4x = 3
\]
\[
x = \frac{3}{4}
\]

#### For the powers of 3:
The powers of 3 give:
\[
3^{-2x} = 3^0
\]

Since \( 3^0 = 1 \), we have:
\[
-2x = 0
\]
\[
x = 0
\]

### Step 4: Verify the solution

The value \( x = 0 \) contradicts the solution from the powers of 2, so the valid solution is \( x = \frac{3}{4} \).

Thus, the solution to the equation is:

\[
x = \frac{3}{4}
\]

Would you like more details or clarifications? Here are five related questions for further practice:

1. Solve \( 5^x = 25^{x+1} \).
2. Solve \( \left( \frac{1}{4} \right)^x = 16^{2x} \).
3. Solve \( 9^{x+1} = 27^{2-x} \).
4. Solve \( 3^{2x} = 9^{x-1} \).
5. Solve \( \left( \frac{1}{2} \right)^x = 4^{x+3} \).

**Tip**: Always check if you can express the bases in terms of smaller prime factors. This makes solving exponential equations easier.

Solve the exponential equation. Express irrational solutions in exact form.
(2/9)^x = 8^(1 - x)

Learn how to solve the exponential equation (2/9)^x = 8^(1 - x) with a detailed step-by-step solution. This problem explores exponent rules and algebraic methods. Suitable for students in Grades 9-12.

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