Question

Find ${\log }_e\left(4.04\right)$
Given: ${\log }_{10}4=0.6021and{\log }_{10}e=0.4343$

Answer 1

Let $y={\log }_{e}x=f\left(x\right)$

Let $\Delta x=0.04x=4$

Now, $y=\log x$

$\Rightarrow \frac{dy}{dx}=\frac{1}{x}$

We know $[\Delta y=\frac{dy}{dx}\Delta xand\Delta y=f(x+\Delta x)-f(x\left)\right]$

$\therefore \Delta y=\frac{dy}{dx}\Delta x=\frac{1}{x}\Delta x$

Putting $x=4$ and $\Delta x=0.4$

$\Delta y=\frac{1}{4}\times 0.04=0.01$

Now

$\Delta y=(x+\Delta x)-f\left(x\right)$ putting all values

$\Rightarrow 0.01={\log }_e(4+0.04)-{\log }_e^4$

$\Rightarrow {\log }_e^{4.04}=0.01+{\log }_e^4$

$=0.01+\frac{{\log }_e^4}{{\log }_{10}^e}$ $[\text{using log property}{\log }_b^a=\frac{{\log }_c^a}{{\log }_c^b}]$

$=0.01+\frac{0.6021}{04343}$ [putting given values]

$[{\log }_e^{4.04}=1.3962]$