If `((9)/(4))^(x)((8)/(27))^(x-1)=(2)/(3)` then the value of x is

Options

A. 1
B. 2
C. 4
D. 3

Explanation

To solve the equation \(\left(\frac{9}{4}\right)^{x} \left(\frac{8}{27}\right)^{x-1} = \frac{2}{3}\), we will follow these steps:

### Step 1: Rewrite the bases in terms of prime factors
We can express the fractions \(\frac{9}{4}\), \(\frac{8}{27}\), and \(\frac{2}{3}\) in terms of their prime factors.

\[ \frac{9}{4} = \frac{3^2}{2^2}, \quad \frac{8}{27} = \frac{2^3}{3^3}, \quad \frac{2}{3} = \frac{2^1}{3^1} \]

### Step 2: Substitute the prime factor forms into the equation
Now we substitute these forms into the equation:

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

### Step 3: Simplify the left side
Using the properties of exponents, we can simplify the left side:

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

This expands to:

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

### Step 4: Combine the exponents
Now we combine the terms with the same bases:

For base \(2\):

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

For base \(3\):

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

So, we rewrite the equation as:

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

### Step 5: Set the exponents equal
Now we can set the exponents equal to each other:

For base \(2\):

\[ x - 3 = 1 \implies x = 4 \]

For base \(3\):

\[ -x + 3 = -1 \implies -x = -4 \implies x = 4 \]

### Conclusion
Both equations give us the same value for \(x\). Therefore, the solution is:

\[ \boxed{4} \]