The approximate value of $\log _{10} 99$ is (Given $\log _{10} \mathrm{e}=0.4343$)
The approximate value of $\log _{10} 99$ is (Given $\log _{10} \mathrm{e}=0.4343$)
- $1.9657$
- $1.9857$
- $1.9957$
- $1.9757$
Solution
Let $f(x)=\log _{10} x=\frac{\log _{e} x}{\log _{e} 10}$
$f^{\prime}(x)=\frac{1}{x \log 10}$
Let $\mathrm{a}=100, \mathrm{~h}=-1$
$\therefore f(a)=\log _{10} 100 \quad=\log _{10} 10^{2}=2$
$f^{\prime}(a)=\frac{1}{100 \times \log _{e} 10}=\frac{1}{100} \log _{10} e=\frac{1}{100} \times 0.4343$
We know that
$\begin{aligned}
f(a-h) & = f(a)+h f^{\prime}(a) \\
& = 2+(-1) \times \frac{1}{100}(0.4343)=2-0.004343=1.995657 \\
& = 1.9957
\end{aligned}$
Asked in: MHT CET 2020 (14 Oct Shift 2)
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