Question

Given the function $y = \log \left( x \right),{\text{ }}0 < x < 10$ , what is the slope of the graph where $x = 5.7?$

Answer 1

In order to solve this question, first we will assume that the $\log$ is taken to base $10$ , then using the logarithmic base change rule i.e., ${\log _b}\left( x \right) = \dfrac{{{{\log }_a}\left( x \right)}}{{{{\log }_a}\left( b \right)}}$ we will change the log base $10$ to log base $e$ .After that we will find the differentiation of the function and we know that $slope = \dfrac{{dy}}{{dx}}$ hence we will get the required slope of the given function.

Complete answer:
The given function is: $y = \log \left( x \right)$
Let us assuming that the log is taken to base $10$
Now we know that
According to the logarithm base change rule:
The base $b$ logarithm of $x$ is base $a$ logarithm of $x$ divided by the base $a$ logarithm of $b$
i.e., ${\log _b}\left( x \right) = \dfrac{{{{\log }_a}\left( x \right)}}{{{{\log }_a}\left( b \right)}}$
So, here we will change the log base $10$ to log base $e$
Therefore, we get
$y = {\log _{10}}\left( x \right) = \dfrac{{{{\log }_e}\left( x \right)}}{{{{\log }_e}\left( {10} \right)}}$
We know that
${\log _e}\left( x \right)$ is also written as $\ln \left( x \right)$
Therefore, from the above equation we get
$y = {\log _{10}}\left( x \right) = \dfrac{{\ln \left( x \right)}}{{\ln \left( {10} \right)}}$
$\Rightarrow y = \dfrac{1}{{\ln \left( {10} \right)}} \cdot \ln \left( x \right){\text{ }} - - - \left( i \right)$
Now we know that,
Slope defines the relationship between the change in y-values with the change in x-values
Mathematically, we can write
$slope = \dfrac{{dy}}{{dx}}$
Therefore, for finding the slope we will have to differentiate the equation $\left( i \right)$
So, on differentiating equation $\left( i \right)$ we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\ln \left( {10} \right)}} \cdot \ln \left( x \right)} \right)$
As $\dfrac{1}{{\ln \left( {10} \right)}}$ is a constant term, so we can take it out from the differentiation.
Therefore, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\ln \left( {10} \right)}} \cdot \dfrac{d}{{dx}}\left( {\ln \left( x \right)} \right)$
As we know that
$\dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}$
Therefore, we have
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\ln \left( {10} \right)}} \cdot \dfrac{1}{x}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{x\ln \left( {10} \right)}}$
Now we know that
$a\ln \left( b \right) = \ln \left( {{b^a}} \right)$
Therefore, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\ln \left( {{{10}^x}} \right)}}$
It is given that $x = 5.7$
On substituting the value, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\ln \left( {{{10}^{5.7}}} \right)}}$
$\ln \left( {{{10}^{5.7}}} \right) \approx 13.124$
Therefore,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{13.124}}$
On dividing, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 0.076$
Hence, the slope is $0.076$

Note:
To solve logarithmic problems, you must know the difference between $\log$ and $\ln$ . $\log$ generally, refers to a logarithm to the base $10$ and known as common logarithm which is represented by ${\log _{10}}\left( x \right)$ . while $\ln$ refers to a logarithm to the base $e$ and known as natural logarithm which is represented by ${\log _e}\left( x \right)$