Using differentials, find the approximate value of `(log)_e 4.04` , it being given that `(log)_(10)4=0. 6021` and `(log)_(10)e=0. 4343` .

Explanation

Let `y=log_ex`

and `x=4`

`x+trianglex=4.04`

`impliestrianglex=0.04`

For `x=4,`

`y=log_e4={log_{10}4}/{log_{10}e}={0.6021}/{0.4343}=1.386368`

Let `dx=trianglex=0.04`

As`y=log_ex`,

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