If the area above the x-axis, bounded by the curves `y = 2^(kx)` and x = 0, and x = 2 is `3/(log_e(2))`, then the value of k is

Options

A. `1//2`
B. 1
C. `-1`
D. 2

Answer

To find the value of \( k \) such that the area above the x-axis, bounded by the curve \( y = 2^{kx} \), \( x = 0 \), and \( x = 2 \) is \( \frac{3}{\log_e(2)} \), we will follow these steps:

### Step 1: Set up the integral for the area
The area \( A \) under the curve \( y = 2^{kx} \) from \( x = 0 \) to \( x = 2 \) can be expressed as:
\[ A = \int_{0}^{2} 2^{kx} \, dx \]

### Step 2: Evaluate the integral
To evaluate the integral, we first rewrite \( 2^{kx} \) using the property of exponents:
\[ 2^{kx} = e^{kx \ln(2)} \]
Now, we can integrate:
\[ A = \int_{0}^{2} e^{kx \ln(2)} \, dx \]
The integral of \( e^{ax} \) is \( \frac{1}{a} e^{ax} \), so we have:
\[ A = \left[ \frac{1}{k \ln(2)} e^{kx \ln(2)} \right]_{0}^{2} \]
Calculating the definite integral:
\[ A = \frac{1}{k \ln(2)} \left( e^{2k \ln(2)} - e^{0} \right) \]
\[ = \frac{1}{k \ln(2)} \left( 2^{2k} - 1 \right) \]

### Step 3: Set the area equal to the given value
According to the problem, this area is equal to \( \frac{3}{\log_e(2)} \):
\[ \frac{1}{k \ln(2)} (2^{2k} - 1) = \frac{3}{\ln(2)} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 2^{2k} - 1 = 3k \]

### Step 5: Rearrange the equation
Rearranging the equation, we have:
\[ 2^{2k} - 3k - 1 = 0 \]

### Step 6: Solve for \( k \)
This equation is transcendental and may not have a simple algebraic solution. We can solve it using numerical methods or graphing techniques. However, we can also check for simple values of \( k \).

Let’s test \( k = 1 \):
\[ 2^{2(1)} - 3(1) - 1 = 4 - 3 - 1 = 0 \]
Thus, \( k = 1 \) is a solution.

### Conclusion
The value of \( k \) is:
\[ \boxed{1} \]