In a reaction the following equation holds good as per chemical kinetics is viewed. `logk=4-4000/(2.303RT)`. The frequency factor is `10^(x)`. What is x.

Explanation

To solve the problem, we start with the given equation from chemical kinetics:

\[ \log k = 4 - \frac{4000}{2.303RT} \]

We need to compare this equation with the general form of the Arrhenius equation, which is:

\[ \log k = \log A - \frac{E_a}{2.303RT} \]

Where:
- \( k \) is the rate constant,
- \( A \) is the frequency factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 1: Identify the components

From the given equation, we can identify the following components:
- The term \( 4 \) corresponds to \( \log A \).
- The term \( 4000 \) corresponds to the activation energy \( E_a \).

### Step 2: Solve for the frequency factor \( A \)

Since \( \log A = 4 \), we can find \( A \) by converting from logarithmic form to exponential form:

\[ A = 10^{\log A} = 10^4 \]

### Step 3: Determine \( x \)

The frequency factor \( A \) can be expressed as \( 10^x \). From our previous calculation, we have:

\[ A = 10^4 \]

This means that:

\[ x = 4 \]

### Final Answer

Thus, the value of \( x \) is:

\[ \boxed{4} \]