The difference of solutions of equation `8^(2x) – (16)8^x + 48 = 0` is

Options

A. `4 sqrt 11`
B. `log_2^3`
C. `1/3 log_2^3`
D. `2 log_2^3`

Answer

To find the difference of the solutions of the equation \( 8^{2x} - 16 \cdot 8^x + 48 = 0 \), we can follow these steps:

### Step 1: Substitute \( t = 8^x \)
We start by substituting \( t = 8^x \). This transforms our equation into a quadratic form:
\[ t^2 - 16t + 48 = 0 \]

### Step 2: Factor the quadratic equation
Next, we will factor the quadratic equation. We need to find two numbers that multiply to \( 48 \) (the constant term) and add up to \( -16 \) (the coefficient of \( t \)). The numbers that satisfy this are \( -12 \) and \( -4 \):
\[ (t - 12)(t - 4) = 0 \]

### Step 3: Solve for \( t \)
Setting each factor to zero gives us the solutions for \( t \):
\[ t - 12 = 0 \quad \Rightarrow \quad t = 12 \]
\[ t - 4 = 0 \quad \Rightarrow \quad t = 4 \]

### Step 4: Substitute back to find \( x \)
Now we substitute back to find \( x \):
1. For \( t = 12 \):
\[ 8^x = 12 \quad \Rightarrow \quad x = \log_8(12) \]
2. For \( t = 4 \):
\[ 8^x = 4 \quad \Rightarrow \quad x = \log_8(4) \]

### Step 5: Calculate the difference of the solutions
We need to find the difference of the solutions \( x_1 = \log_8(12) \) and \( x_2 = \log_8(4) \):
\[ \text{Difference} = x_1 - x_2 = \log_8(12) - \log_8(4) \]
Using the properties of logarithms, we can combine these:
\[ \text{Difference} = \log_8\left(\frac{12}{4}\right) = \log_8(3) \]

### Final Result
Thus, the difference of the solutions of the equation is:
\[ \log_8(3) \]

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