If `(0.2)^((2x-3)/(x-2))gt 5` then complete solution set of values of x is

Options

A. `(5//3, 2)`
B. `(3//5, 2)`
C. `(-oo, -(5)/(3))uu(2,oo)`
D. `((5)/(3),oo)`

Explanation

To solve the inequality \( (0.2)^{\frac{2x-3}{x-2}} > 5 \), we will follow these steps:

### Step 1: Rewrite the base
First, we can rewrite \( 0.2 \) as \( \frac{1}{5} \):
\[ (0.2)^{\frac{2x-3}{x-2}} = \left(\frac{1}{5}\right)^{\frac{2x-3}{x-2}} \]

### Step 2: Change the inequality
This can be rewritten as:
\[ \left(\frac{1}{5}\right)^{\frac{2x-3}{x-2}} > 5 \]
Taking the reciprocal, we get:
\[ \left(5\right)^{-\frac{2x-3}{x-2}} > 5 \]

### Step 3: Rewrite the inequality
Now we can express \( 5 \) as \( 5^1 \):
\[ 5^{-\frac{2x-3}{x-2}} > 5^1 \]

### Step 4: Set the exponents
Since the bases are the same, we can set the exponents:
\[ -\frac{2x-3}{x-2} > 1 \]

### Step 5: Multiply both sides by \(-1\)
When we multiply both sides by \(-1\), we must reverse the inequality:
\[ \frac{2x-3}{x-2} < -1 \]

### Step 6: Rearranging the inequality
Rearranging gives:
\[ 2x - 3 < - (x - 2) \]
\[ 2x - 3 < -x + 2 \]

### Step 7: Combine like terms
Adding \( x \) to both sides:
\[ 2x + x < 2 + 3 \]
\[ 3x < 5 \]

### Step 8: Solve for \( x \)
Dividing both sides by 3:
\[ x < \frac{5}{3} \]

### Step 9: Find the critical point
Now, we also need to consider the denominator in the original expression \( x - 2 \neq 0 \), which gives \( x \neq 2 \).

### Step 10: Combine the results
Thus, the solution set is:
\[ x < \frac{5}{3} \quad \text{and} \quad x \neq 2 \]

### Final Solution
The complete solution set of values of \( x \) is:
\[ (-\infty, \frac{5}{3}) \]