Question

Find the appropriate value of ${\log }_{10}$ $\left(1016\right)$, given ${\log }_{10}e=0.4343$

Answer 1

$Lo{g}_{10}\left(1016\right)=?{\log }_{10}e=0.4343$

$Letf\left(x\right)={\log }_{10}x\to \left(i\right)f\left(x\right)=\frac{{\log }_{e}x}{{\log }_{e}10}\to \left(ii\right)={\log }_{e}x\cdot {\log }_{10}e=0.4343{\log }_{e}x$

On differentiating w.r.t $x$, we get

$f^{\prime }\left(x\right)=\frac{0.4343}{x}$

Since,

$x=1016=1000+16$

$Leth=16,a=1000f\left(a\right)=f\left(1000\right)={\log }_{10}1000=3{\log }_{10}10=3\times 1=3$

$f^{\prime }\left(a\right)=f^{\prime }\left(1000\right)=\frac{0.4343}{1000}=0.0004343f\left(a+h\right)\approx f\left(a\right)+h{f}^{\prime }\left(a\right)\approx 3+1610.0004343k\approx 3+0.009488$

$\approx 3.0069488$

Hence, this is the answer.