Question

If a rough approximation for $\ln (5)$ is $1.609$ how do you use this approximation and differentials to approximate $\ln \left( {\dfrac{{128}}{{25}}} \right)$ ?

Answer 1

In is the natural logarithm. It is log to the base of $e$ . $e$ is an irrational and transcendental number the first few digits of which are: $2.71828182459....$ in higher mathematics the natural logarithm is the log that is usually used.
Differentiation of $\ln x = \dfrac{1}{x}$ .

Complete step by step solution:
To approximate $\ln \left( {\dfrac{{128}}{{25}}} \right)$ using linear approximation and differential
We need a number near $\dfrac{{128}}{{25}}$ whose $\ln$ we know.
We get, $\dfrac{{128}}{{25}}$ is somewhat near $\dfrac{{125}}{{25}}$
And $\dfrac{{125}}{{25}} = 5$ whose $\ln$ we were given in the question.
As, the difference between $\ln \left( {\dfrac{{128}}{{25}}} \right)$ and $\ln (5)$ is approximately equal to the differential of $y = In\,x$
Differentiate $y$ w.r.t $x$ . We get,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\ln \,x = \dfrac{1}{x}$
$\Rightarrow dy = \dfrac{1}{x}dx$
To approximate near $5$ , we will use $dy = \dfrac{1}{5}dx = \dfrac{1}{5}(x - 5)$
With $x = \dfrac{{128}}{{25}}$
$\Rightarrow x - 5 = \dfrac{{128}}{{25}} - 5 = \dfrac{3}{{25}}$
Multiply $\dfrac{4}{4}$ . We get,
$\Rightarrow \dfrac{3}{{25}} \times 4 = \dfrac{{12}}{{100}} = 0.12$
And $dy = \dfrac{1}{2}(0.12) = 0.024$
We can write,
$\ln \left( {\dfrac{{128}}{{25}}} \right) = \ln \left( {\dfrac{{125}}{{25}}} \right) + \Delta y$
We can write $\Delta y$ as $dy$ . So,
$\ln \left( {\dfrac{{128}}{{25}}} \right) \approx \ln \left( {\dfrac{{125}}{{25}}} \right) + dy \approx 1.609 + 0.024 = 1.633$

Hence, $\ln \left( {\dfrac{{128}}{{25}}} \right) = 1.633$

Note: Natural logarithm is used in science, engineering, and physics fields. It is used in calculation in the age of ancient by using carbon dating in which we calculate the number of $C - 14$ . It is used in calculation of radioactivity as well as in determining the rate of reaction of process in labs and nature.