If f(x) = `2^(x) + 1 ` , then `f^(-1)(7)` =

Options

A. 2.4
B. 2.6
C. 2.8
D. 3

Explanation

To find \( f^{-1}(7) \) for the function \( f(x) = 2^x + 1 \), we will follow these steps:

### Step 1: Set up the equation
We start by setting \( f(z) = 7 \), where \( z \) is the value we need to find such that \( f(z) = 2^z + 1 \).

### Step 2: Solve for \( z \)
From the equation \( f(z) = 7 \), we have:
\[ 2^z + 1 = 7 \]
Subtracting 1 from both sides gives:
\[ 2^z = 6 \]

### Step 3: Take the logarithm
To solve for \( z \), we take the logarithm (base 2) of both sides:
\[ \log_2(2^z) = \log_2(6) \]
Using the property of logarithms, \( \log_a(b^c) = c \cdot \log_a(b) \), we simplify the left side:
\[ z = \log_2(6) \]

### Step 4: Express \( \log_2(6) \) in terms of natural logarithms
Using the change of base formula for logarithms, we can express \( \log_2(6) \) as:
\[ z = \frac{\log(6)}{\log(2)} \]

### Conclusion
Thus, we have:
\[ f^{-1}(7) = \frac{\log(6)}{\log(2)} \]

### Final Answer
The value of \( f^{-1}(7) \) is \( \frac{\log(6)}{\log(2)} \).

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