Question

If ${\log }_e\left(5\right)=1.609$, then the approximate value of ${\log }_e\left(5.1\right)$ is

• A. $1.611$
• B. $1.701$
• C. $1.809$
• D. $1.629$

Answer 1

The correct option is D

$1.629$

Explanation for the correct option

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

Let $x_0=5$, $dx=0.1$

Using the formula to find value of approximate function

$\mathbf{f}\left(x+\mathrm{dx}\right)\mathbf{=}\mathbf{f}\left(x_0\right)\mathbf{+}\mathbf{dx}\mathbf{\times }\mathbf{f}\mathbf{\text{'}}\left(x_0\right)$

where $f\text{'}\left(x_0\right)=\frac{d}{dx}\left(f\left(x\right)\right)$ at $x=x_0$

Differentiating $f\left(x\right)$ with respect to $x$ we get,

$f\text{'}\left(x\right)=\frac{d}{dx}\left[{\log }_e\left(x\right)\right]=\frac{1}{x}$

Substitute $x=5$ we get

$f\text{'}\left(x_0\right)=\frac{1}{5}=0.2$

$f\left(x_0\right)={\log }_e\left(5\right)=1.609$

$dx=0.1$

Substituting all the required values in formula for approximate value we get

$\log \left(5+0.1\right)=1.609+0.1\times 0.2$

$\Rightarrow$ $\log \left(5.1\right)=1.609+0.02$

$\Rightarrow$ $\log \left(5.1\right)=1.629$

Hence, the approximate value of $\log 5.1$ is $1.629$

Hence, option(D) i.e. $1.629$ is the correct answer.