1.

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

Answer» 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`
`=3+0.0021`
`=3.0021`
`:. log_10(1005) = 3.0021`


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