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)(logx)=(0.4343)logx`
`:.f'(x)=(0.4343).("d")/("dx")(logx)=(0.4343)/(x)`
Take `a=1000" and "h=16`. Then
`f(a)=f(1000)=log_(10)1000=log_(10)10^(3)=3`
`f'(a)=f'(1000)=(0.4343)/(1000)`
The formula for approximation is
`f(a+h)=f(a)+hf'(a)`
`:.log_(10)1016=f(1016)=f(1000+16)`
`=f(1000)+16.f'(1000)`
`=3+16xx(0.4343)/(1000)`
`=3+0.0069488=3.006949`
`:. log_(10)1016=3.006949`.