Question

The approximate value of $\log_{10} 998$ is (given that $\log_{10}e=0.4343$)

• A. 3.0008686
• B. 1.9991314
• C. 2.0008686
• D. 2.9991314

Hint:

Logic Tip: Be careful with bases! The derivative of $\ln(x)$ is $1/x$, but the derivative of $\log_{10}(x)$ involves the conversion factor $\log_{10}(e)$ or $1/\ln(10)$. Forgetting this factor is a common trap in approximation problems.

Answer 1

The Correct Option is D

Concept:
The linear approximation (or differential approximation) of a function $f(x)$ near a point $a$ is given by: $$f(a + h) \approx f(a) + h \cdot f'(a)$$ where $a$ is a point where the function value is known easily, and $h$ is a small increment (positive or negative).
Step 1: Define the function and identify a, h.
Let $f(x) = \log_{10} x$. We want to find $f(998)$. We can choose $a = 1000$ because $\log_{10} 1000$ is easy to calculate. Thus, $x = a + h \implies 998 = 1000 + (-2)$. Here, $a = 1000$ and $h = -2$.
Step 2: Calculate f(a) and f'(x).
Find $f(a)$: $$f(1000) = \log_{10}(1000) = \log_{10}(10^3) = 3$$ Next, find the derivative $f'(x)$. Using the change of base formula, $\log_{10} x = \frac{\ln x}{\ln 10} = (\log_{10} e) \cdot \ln x$. $$f'(x) = \frac{d}{dx}\left( (\log_{10} e) \ln x \right)$$ $$f'(x) = (\log_{10} e) \cdot \frac{1}{x}$$ Given $\log_{10} e = 0.4343$, we have: $$f'(x) = \frac{0.4343}{x}$$
Step 3: Evaluate f'(a) and apply the approximation formula.
Evaluate the derivative at $a = 1000$: $$f'(1000) = \frac{0.4343}{1000} = 0.0004343$$ Now, substitute $f(a)$, $h$, and $f'(a)$ into the approximation formula: $$f(a + h) \approx f(a) + h \cdot f'(a)$$ $$\log_{10}(998) \approx 3 + (-2)(0.0004343)$$ $$\log_{10}(998) \approx 3 - 0.0008686$$ $$\log_{10}(998) \approx 2.9991314$$