`l_(1)=int2^(x)dx=p(x)+c_(1)andl_(2)=int((1)/(2))^(x)dx=m(x)+c_(1)` then p(x)-m(x) is equal to

Options

A. `{log_(e)(2)}(2^(x)-2^(-x))`
B. `{log_(e)(2)}(2^(x)+2^(-x))`
C. `(1)/(log_(e)2)(2^(x)+2^(-x))`
D. `(log_(2)e)(2^(x)+2^(-x))`

Answer

To solve the problem, we need to find the difference \( p(x) - m(x) \) where:

- \( L_1 = \int 2^x \, dx = p(x) + c_1 \)
- \( L_2 = \int \frac{1}{2^x} \, dx = m(x) + c_1 \)

### Step 1: Find \( p(x) \)

We start with the integral \( L_1 = \int 2^x \, dx \).

Using the formula for the integral of an exponential function, we have:

\[ \int a^x \, dx = \frac{a^x}{\ln a} + C \]

For \( a = 2 \):

\[ L_1 = \int 2^x \, dx = \frac{2^x}{\ln 2} + C_1 \]

Thus, we can express \( p(x) \) as:

\[ p(x) = \frac{2^x}{\ln 2} \]

### Step 2: Find \( m(x) \)

Next, we find \( L_2 = \int \frac{1}{2^x} \, dx \).

We can rewrite \( \frac{1}{2^x} \) as \( 2^{-x} \):

\[ L_2 = \int 2^{-x} \, dx \]

Using the same formula for the integral of an exponential function:

\[ L_2 = \int 2^{-x} \, dx = \frac{2^{-x}}{\ln 2} + C_1 \]

Thus, we can express \( m(x) \) as:

\[ m(x) = \frac{2^{-x}}{\ln 2} \]

### Step 3: Calculate \( p(x) - m(x) \)

Now, we need to find \( p(x) - m(x) \):

\[ p(x) - m(x) = \frac{2^x}{\ln 2} - \frac{2^{-x}}{\ln 2} \]

Factoring out \( \frac{1}{\ln 2} \):

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

### Final Result

Thus, the final expression for \( p(x) - m(x) \) is:

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