The number of solutions of `log_(x)3. = 2x-3` is,

Options

A. 3
B. 1
C. 2
D. 0

Answer

To solve the equation \( \log_x 3 = 2x - 3 \), we can follow these steps:

### Step 1: Rewrite the logarithmic equation
Using the change of base formula for logarithms, we can rewrite the equation as:
\[ \frac{\log 3}{\log x} = 2x - 3 \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:
\[ \log 3 = (2x - 3) \log x \]

### Step 3: Define the functions
Let:
\[ f(x) = \log 3 \]
and
\[ g(x) = (2x - 3) \log x \]
We need to find the number of intersections between the two functions \( f(x) \) and \( g(x) \).

### Step 4: Analyze the function \( g(x) \)
The function \( g(x) \) is defined only for \( x > 0 \) and \( x \neq 1 \) (since \( \log x \) is undefined at \( x = 1 \)).

- As \( x \to 0^+ \), \( \log x \to -\infty \) and thus \( g(x) \to 0 \).
- As \( x \to 1^- \), \( g(x) \to -\infty \).
- As \( x \to 1^+ \), \( g(x) \to +\infty \).
- As \( x \to \infty \), \( g(x) \) will also approach \( +\infty \) since \( 2x - 3 \) becomes positive and \( \log x \) increases.

### Step 5: Find critical points
To find the critical points of \( g(x) \), we can differentiate \( g(x) \):
\[ g'(x) = 2 \log x + (2x - 3) \cdot \frac{1}{x} \]
Setting \( g'(x) = 0 \) will help us find the local maxima and minima.

### Step 6: Determine the number of intersections
By analyzing the behavior of \( f(x) \) and \( g(x) \):
- \( f(x) = \log 3 \) is a constant line.
- \( g(x) \) starts from \( 0 \) as \( x \to 0^+ \), goes to \( -\infty \) as \( x \to 1^- \), goes to \( +\infty \) as \( x \to 1^+ \), and then increases towards \( +\infty \).

Since \( g(x) \) crosses the line \( y = \log 3 \) twice (once before \( x = 1 \) and once after \( x = 1 \)), we conclude that there are two solutions to the equation.

### Final Answer
The number of solutions of the equation \( \log_x 3 = 2x - 3 \) is **2**.

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