Question

If $log3 = 1.0986$ then the approximate value of $log\left( {9.01} \right)$ is
A) $2.1983$
B) $2.1893$
C) $2.1389$
D) $2.1789$

Answer 1

The derivative of a logarithmic function is the reciprocal of the argument. As always, the chain rule tells us to also multiply by the derivative of the argument. So if $f\left( x \right) = \log \left( u \right)$ then
$f'\left( x \right) = \dfrac{1}{u}.u'$; $\dfrac{d}{{dx}}(\log x) = \dfrac{1}{x}$; $\dfrac{d}{{dx}}\left( {lo{g_b}x} \right) = \dfrac{1}{{\left( {\log b} \right)x}}$
Basic Idea: the derivative of a logarithmic function is that of the reciprocal of the things inside.
Using the properties of logarithms will sometimes make the differentiation process easier.
The derivative of a logarithmic function has been proved by using first principles.

Complete step-by-step answer:
It is given within the problem $log3 = 1.0986$ and finds the approximate value of $log\left( {9.01} \right)$.
For this, we will assume that $y = logx$
When differentiating the equation, we get then, $\vartriangle x = \dfrac{{dy}}{{dx}} = \dfrac{1}{x}$
And when $x{\text{ = }}9$so we get $f'(x) = \vartriangle x = \dfrac{1}{9}$
Now, we will assume that $x{\text{ = }}9$,$\vartriangle x{\text{ = }}0.01$and $f(x + \vartriangle x) = log\left( {9.01} \right)$
Now, for finding the approximate value of $f(x + \vartriangle x) = log\left( {9.01} \right)$.

$$\Rightarrow f(x){\text{ }} = {\text{ }}log\left( {{\text{9 }} + {\text{ 0}}{\text{.01}}} \right) \\\ \Rightarrow f(x){\text{ }} = {\text{ }}log\left( {{\text{x }} + {\text{ }}\vartriangle {\text{x}}} \right){\text{ }}$$

Now on differentiating the equation $f(x) = log\left( {9.01} \right)$ with respect to $x$
Then the derivative of f(x) is given by:
$\Rightarrow {f'}(x){\text{ }} = {\text{ }}\mathop {\lim }\limits_{\vartriangle x \to 0} \dfrac{{f(x + \vartriangle x) - f(x)}}{{\vartriangle x}}$
We write this as:
$\Rightarrow f(x + \vartriangle x) = \vartriangle x{f'}(x) + f(x)$
$\Rightarrow f(x + \vartriangle x) = 0.01{f'}(9) + f(9)$
Substituting the value of x =9 in the above equation.
$\Rightarrow f(x + \vartriangle x) = 0.01{f'}(9) + \log 9$
Replacing 9 with 32 as nine is equal to square three.
$\Rightarrow f(x + \vartriangle x) = 0.01{f'}(9) + \log {3^2}$
Substituting the value of $f'(x) = \vartriangle x = \dfrac{1}{9}$ in the above equation.
$\Rightarrow f(x + \vartriangle x) = 0.01\left( {\dfrac{1}{9}} \right) + 2\log 3$
Here, we will substitute the value of $log3 = 1.0986$ in the above equation.
$\Rightarrow f(x + \vartriangle x) = \dfrac{{0.01}}{9} + 2.1972$
So the final answer is :
$\Rightarrow f(x + \vartriangle x) = 0.0011 + 2.1972 = 2.1983$
So, the approximate value of $log\left( {9.01} \right)$ is $2.1983$

Therefore, the option (A) is the correct answer.

Note: In this question the value of $log\left( {9.01} \right)$ has to be calculated. Unfortunately, we can only use the logarithm laws to help us in a limited number of logarithm differentiation question types. Most often, we like to seek out the derivative of a logarithm of some function of x. Differentiate the logarithmic functions.