Using differentials, find the approximate value of `(log)_(10)10. 1` , it being given that `(log)_(10)e=0. 4343` .

Explanation

Let `y=log_{10}x`

and `x=10`

`x+trianglex=10.1`

`impliestrianglex=0.1`

For `x=10,`

`y=log_{10}10=1`

Let `dx=trianglex=0.1`

As`y=log_{10}x`,

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