Find the approximate value of `log_10(1016)` given `log_10 e =0.4343`

Explanation

Let `f(x)=log_(10)x=(log_(e)x)/(log_(e)10)`
`=(log_(10)e)(log x)`
`=(0.4343)*log x`
`:.f'(x)=0.4343*(d)/(dx)(log x)`
`=(0.4343)/(x)`
Take `a=1000, h = 16` then
`f(a)=f(1000)=log_(10)1000`
`=log_(10)10^(3)=3*log_(10)10=3`
`f'(a)=f'(1000)=(0.4343)/(1000)=0.0004343`
The formula for approximation is
` f(a+h) underset(.)overset(.)= f(a)+h*f'(a)`
`:.log_(10)1016 underset(.)overset(.)= f(1016)underset(.)overset(.)= f'(1000+16)`
`underset(.)overset(.)= f(1000)+16 * f'(1000)`
`underset(.)overset(.)= 3+16xx(0.0004343)`
`underset(.)overset(.)= 3+0.0069488 underset(.)overset(.)= 3*006949`
`:.log_(10)1016=3*006949`