Use differential to approximate `log(9.01).` (Given, `log3=1.0986`)

Answer

To approximate \( \log(9.01) \) using differentials, we can follow these steps:

### Step 1: Identify the function and the point of approximation
Let \( f(x) = \log(x) \). We want to approximate \( \log(9.01) \). We can express \( 9.01 \) as \( 9 + 0.01 \), where \( x = 9 \) and \( \Delta x = 0.01 \).

### Step 2: Use the differential approximation formula
The differential approximation formula is given by:
\[ f(x + \Delta x) \approx f(x) + f'(x) \Delta x \]
In our case, we need to find \( f(9) \) and \( f'(9) \).

### Step 3: Calculate \( f(9) \)
Using the properties of logarithms:
\[ f(9) = \log(9) = \log(3^2) = 2 \log(3) \]
Given \( \log(3) = 1.0986 \):
\[ f(9) = 2 \times 1.0986 = 2.1972 \]

### Step 4: Calculate \( f'(x) \)
The derivative of \( f(x) = \log(x) \) is:
\[ f'(x) = \frac{1}{x} \]
Thus,
\[ f'(9) = \frac{1}{9} \]

### Step 5: Substitute values into the approximation formula
Now we can substitute \( f(9) \), \( f'(9) \), and \( \Delta x \) into the approximation formula:
\[ f(9 + 0.01) \approx f(9) + f'(9) \Delta x \]
\[ \log(9.01) \approx 2.1972 + \left(\frac{1}{9}\right)(0.01) \]
Calculating \( \frac{1}{9} \times 0.01 \):
\[ \frac{1}{9} \times 0.01 = \frac{0.01}{9} \approx 0.0011 \]

### Step 6: Final calculation
Now, add this to \( f(9) \):
\[ \log(9.01) \approx 2.1972 + 0.0011 = 2.1983 \]

### Conclusion
Thus, the approximate value of \( \log(9.01) \) is:
\[ \log(9.01) \approx 2.1983 \]

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