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

Hint: 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 \[{\log _e}\left( x \right)\]