Find the approximate value of `(log)_(10)1005` , given that `(log)_(10)e=0. 4343`

Explanation

Here, we will use the following rule,

`f(x+Deltax) = f(x) + f\'(x)Deltax`

Here, `f(x+Deltax) = log_10 1005, f(x) = log_10 x, x = 1000, Deltax = 5`

`:. log_10(x+Deltax) = log_10 x+d/dx(log_10x)Deltax`

`:. log(x+Deltax) = log_10x+d/dx(log_e x/log_e 10)Deltax`

`=>log_10(1005) = log_10 (1000)+1/(log_e 10)(1/x)(5)`

`= 3+log_10 e(1/1000)**5`
`=3+0.4343**1/1000**5`

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