The approximate value of log _{10} 1002 is (G | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We use the linear approximation formula: $f(a+h) \approx f(a) + h f'(a)$.Let $f(x) = \log_{10} x$.Then $f'(x) = \frac{1}{x \ln 10} = \frac{\log_{1

Vedclass · Accepted Answer

$3.0009$

Vedclass · Answer

(B) We use the linear approximation formula: $f(a+h) \approx f(a) + h f'(a)$.Let $f(x) = \log_{10} x$.Then $f'(x) = \frac{1}{x \ln 10} = \frac{\log_{10} e}{x}$.Given $a = 1000$ and $h = 2$,we have:$f(1002) \approx f(1000) + 2 f'(1000)$.$f(1000) = \log_{10} 1000 = 3$.$f'(1000) = \frac{\log_{10} e}{1000} = \frac{0.4343}{1000} = 0.0004343$.Therefore,$\log_{10} 1002 \approx 3 + 2(0.0004343)$.$\log_{10} 1002 \approx 3 + 0.0008686 = 3.0008686$.Rounding to four decimal places,we get $3.0009$.

The approximate value of $\log _{10} 1002$ is (Given $\log _{10} e = 0.4343$)

Similar Questions

Find the approximate value of $(32.15)^{\frac{1}{5}}$.

Find the approximate value of $(401)^{\frac{1}{2}}$.

If $y = x^3 + 5$ and $x$ changes from $3$ to $2.99$,then the approximate change in $y$ is

The circumference of a circle is measured as $56 \text{ cm}$ with an error of $0.02 \text{ cm}$. The percentage error in its area is

The diameter of a sphere is measured as $42 \text{ cm}$. If there is an error of $1/77 \text{ cm}$ in measuring it, then the error involved in the volume of that sphere (in cubic centimeters) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module