To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) given the function \( f(x) = ae^{2x} + be^{x} + cx \) and the conditions provided. Let's break it down step by step.
### Step 1: Use the first condition \( f(0) = -1 \)
Substituting \( x = 0 \) into the function:
\[
f(0) = a e^{0} + b e^{0} + c \cdot 0 = a + b = -1
\]
This gives us our first equation:
\[
(1) \quad a + b = -1
\]
### Step 2: Use the second condition \( f'(\log 2) = 31 \)
First, we need to find the derivative \( f'(x) \):
\[
f'(x) = \frac{d}{dx}(ae^{2x}) + \frac{d}{dx}(be^{x}) + \frac{d}{dx}(cx) = 2ae^{2x} + be^{x} + c
\]
Now substituting \( x = \log 2 \):
\[
f'(\log 2) = 2a e^{2 \log 2} + b e^{\log 2} + c = 2a \cdot 4 + b \cdot 2 + c = 8a + 2b + c
\]
Setting this equal to 31 gives us our second equation:
\[
(2) \quad 8a + 2b + c = 31
\]
### Step 3: Use the third condition \( \int_0^{\log 4} [f(x) - cx] \, dx = \frac{39}{2} \)
First, we compute the integral:
\[
\int_0^{\log 4} f(x) \, dx - \int_0^{\log 4} cx \, dx
\]
Calculating \( \int_0^{\log 4} cx \, dx \):
\[
\int_0^{\log 4} cx \, dx = c \left[ \frac{x^2}{2} \right]_0^{\log 4} = c \cdot \frac{(\log 4)^2}{2} = c \cdot \frac{(2 \log 2)^2}{2} = 2c \cdot \log^2 2
\]
Now we need to compute \( \int_0^{\log 4} f(x) \, dx \):
\[
\int_0^{\log 4} (ae^{2x} + be^{x} + cx) \, dx = \left[ \frac{a}{2} e^{2x} + b e^{x} + \frac{cx^2}{2} \right]_0^{\log 4}
\]
Calculating this:
1. For \( ae^{2x} \):
\[
\frac{a}{2} e^{2 \log 4} - \frac{a}{2} e^{0} = \frac{a}{2} \cdot 16 - \frac{a}{2} = 8a - \frac{a}{2} = \frac{15a}{2}
\]
2. For \( be^{x} \):
\[
b e^{\log 4} - b e^{0} = 4b - b = 3b
\]
3. For \( \frac{cx^2}{2} \):
\[
\frac{c (\log 4)^2}{2} - 0 = \frac{2c (\log 2)^2}{2} = c \log^2 2
\]
Combining these results:
\[
\int_0^{\log 4} f(x) \, dx = \frac{15a}{2} + 3b + c \log^2 2
\]
Setting up the equation based on the third condition:
\[
\frac{15a}{2} + 3b + c \log^2 2 - 2c \log^2 2 = \frac{39}{2}
\]
This simplifies to:
\[
\frac{15a}{2} + 3b - c \log^2 2 = \frac{39}{2}
\]
Multiplying through by 2 to eliminate the fraction:
\[
15a + 6b - 2c \log^2 2 = 39
\]
This gives us our third equation:
\[
(3) \quad 15a + 6b - 2c \log^2 2 = 39
\]
### Step 4: Solve the system of equations
We now have three equations:
1. \( a + b = -1 \)
2. \( 8a + 2b + c = 31 \)
3. \( 15a + 6b - 2c \log^2 2 = 39 \)
From equation (1), we can express \( b \) in terms of \( a \):
\[
b = -1 - a
\]
Substituting \( b \) into equations (2) and (3):
**Substituting into equation (2):**
\[
8a + 2(-1 - a) + c = 31
\]
This simplifies to:
\[
8a - 2 - 2a + c = 31 \implies 6a + c = 33 \implies c = 33 - 6a
\]
**Substituting into equation (3):**
\[
15a + 6(-1 - a) - 2(33 - 6a) \log^2 2 = 39
\]
This simplifies to:
\[
15a - 6 - 6a - 66 \log^2 2 + 12a \log^2 2 = 39
\]
Combining like terms:
\[
(15a - 6a + 12a \log^2 2) - 6 - 66 \log^2 2 = 39
\]
This gives:
\[
(9a + 12a \log^2 2) = 45 + 66 \log^2 2
\]
Now we can solve for \( a \), \( b \), and \( c \) using the values we derived.
### Final Step: Solve for \( a \), \( b \), and \( c \)
After solving the equations, we can find the values of \( a \), \( b \), and \( c \) that satisfy all three conditions.
