Find the approximate value of _ 10 (1016) given that _ 10 e=0.434… - Vidyadip

Question

Find the approximate value of $\log _{10}(1016)$ given that $\log _{10} e=0.4343$.

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Answer

Let $f ( x )=\log _{10} x =\frac{\log _e x}{\log _e 10}$
$= (log_{10}e)(log_ex)$
$= (0.4343) \log x$
$\therefore f ^{\prime}( x )=\frac{0.4343}{x}$
$x = 1016 = 1000 + 16 = a + h$
Here$, a = 1000$ and $h = 16$
$f(a) = f(1000)$
$= log_{10}(1000)$
$= log_{10}(10)^3$
$= 3log_{10} 10 ...[\because log10 m^n = n log_{10} m]$
$= 3$
$f^{\prime}(a)=f^{\prime}(1000)=\frac{0.4343}{1000}=0.0004343$
$f(a+h) \approx f(a)+h f^{\prime}(a)$
$\log _{10}(1016) \approx 3+16(0.0004343)$
$\approx 3+0.0069488$
$\log _{10}(1016) \approx 3.006949$

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