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

Explanation

Let `y=log_{10}x`

and `x=1000`

`x+trianglex=1005`

`impliestrianglex=5`

For `x=1000,`

`y=log_{10}1000=3`

Let `dx=trianglex=5`

As`y=log_{10}x=frac{log_ex}{log_e{10}}`,

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