If loge 4 = 1.3868, then loge 4.01 =

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1.3893 \text{ Consider the function } y = f\left( x \right) = \log_e x . \text { Let }: x = 4 x + ∆ x = 4 . 01 \Rightarrow ∆ x = 0 . 01 \text { For }x = 4, y = l {og}_e 4 = 1 . 3868 y = \log_e x \Rightarrow dy/dx = 1/x \Rightarrow \left( dy/dx \right)_{x = 4} = $\frac{1}{4}$ \Rightarrow ∆ y = dy = dy/dxdx = $\frac{1}{4}$ $\times$ 0 . 01 = 0 . 0025 \therefore \log_e 4 . 01 = y + ∆ y = 1 . 3893

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