
Why does $$\ln a - \ln b = \ln \left( {\dfrac{a}{b}} \right)$$?
Answer
479.1k+ views
Hint: Here in this question, given one of the properties of logarithm function we need to prove that. For this, we need to consider the inverses of exponential functions then by using some different Basic Properties of logarithmic and exponential function we get the required solution.
Complete step by step answer:
The function from positive real numbers to real numbers to real numbers is defined as $${\log _b}:{R^ + } \to R \Rightarrow {\log _b}\left( x \right) = y$$, if $${b^y} = x$$, is called logarithmic function or the logarithm function is the inverse form of exponential function.
Now consider the given question:
We need to prove $$\ln a - \ln b = \ln \left( {\dfrac{a}{b}} \right)$$.
Basically, logarithms can be divided into two types:
Logarithms of the base ‘10’ are called common logarithms. And Logarithms of the base ‘$$e$$’ are called natural logarithms.
It does not matter what base we use, providing the same base is used for all logarithms, here we are using base ‘$$e$$’.
Let us define $$A$$, $$B$$, $$C$$ as follows:
$$A = \ln a\, \Leftrightarrow \,a = {e^A}$$ --- (1)
$$B = \ln b\, \Leftrightarrow \,b = {e^B}$$ --- (2)
$$C = \ln \left( {\dfrac{a}{b}} \right) \Leftrightarrow \,\,\dfrac{a}{b} = {e^C}$$ --- (3)
From the last definition we have:
$$ \Rightarrow \,\,\,\,{e^C} = \dfrac{a}{b}$$
From equation (1) and (2), then we have
$$ \Rightarrow \,\,\,\,{e^C} = \dfrac{{{e^A}}}{{{e^B}}}$$
By using the law of indices $${a^{ - n}} = \dfrac{1}{{{a^n}}}$$ and $${a^m} \cdot {a^n} = {a^{m + n}}$$, then
$$ \Rightarrow \,\,\,\,{e^C} = \left( {{e^A}} \right) \cdot \left( {{e^{ - B}}} \right)$$
$$ \Rightarrow \,\,\,\,{e^C} = {e^{A - B}}$$
Here, the exponential is a $$1:1$$ monotonic continuous function, then we have
$$ \Rightarrow \,\,C = A - B$$
From equation (1), (2) and (3)
$$\therefore \,\,\,\,\,\ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b$$
Hence, the required solution.
Note:
If the function contains the log term, then the function is known as logarithmic function. We have two types of logarithms namely common logarithm and natural logarithm. Since it involves the arithmetic operations, we have a standard logarithmic property for the arithmetic operations. By using the properties, we can solve these types of questions.
There are other some basic logarithms properties:
1. product rule :- $$\log \left( {mn} \right) = \log m + \log n$$
2. Quotient rule :- $$\log \left( {\dfrac{m}{n}} \right) = \log m - \log n$$
3. Power rule :- $$\log \left( {{m^n}} \right) = n.\log m$$
Complete step by step answer:
The function from positive real numbers to real numbers to real numbers is defined as $${\log _b}:{R^ + } \to R \Rightarrow {\log _b}\left( x \right) = y$$, if $${b^y} = x$$, is called logarithmic function or the logarithm function is the inverse form of exponential function.
Now consider the given question:
We need to prove $$\ln a - \ln b = \ln \left( {\dfrac{a}{b}} \right)$$.
Basically, logarithms can be divided into two types:
Logarithms of the base ‘10’ are called common logarithms. And Logarithms of the base ‘$$e$$’ are called natural logarithms.
It does not matter what base we use, providing the same base is used for all logarithms, here we are using base ‘$$e$$’.
Let us define $$A$$, $$B$$, $$C$$ as follows:
$$A = \ln a\, \Leftrightarrow \,a = {e^A}$$ --- (1)
$$B = \ln b\, \Leftrightarrow \,b = {e^B}$$ --- (2)
$$C = \ln \left( {\dfrac{a}{b}} \right) \Leftrightarrow \,\,\dfrac{a}{b} = {e^C}$$ --- (3)
From the last definition we have:
$$ \Rightarrow \,\,\,\,{e^C} = \dfrac{a}{b}$$
From equation (1) and (2), then we have
$$ \Rightarrow \,\,\,\,{e^C} = \dfrac{{{e^A}}}{{{e^B}}}$$
By using the law of indices $${a^{ - n}} = \dfrac{1}{{{a^n}}}$$ and $${a^m} \cdot {a^n} = {a^{m + n}}$$, then
$$ \Rightarrow \,\,\,\,{e^C} = \left( {{e^A}} \right) \cdot \left( {{e^{ - B}}} \right)$$
$$ \Rightarrow \,\,\,\,{e^C} = {e^{A - B}}$$
Here, the exponential is a $$1:1$$ monotonic continuous function, then we have
$$ \Rightarrow \,\,C = A - B$$
From equation (1), (2) and (3)
$$\therefore \,\,\,\,\,\ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b$$
Hence, the required solution.
Note:
If the function contains the log term, then the function is known as logarithmic function. We have two types of logarithms namely common logarithm and natural logarithm. Since it involves the arithmetic operations, we have a standard logarithmic property for the arithmetic operations. By using the properties, we can solve these types of questions.
There are other some basic logarithms properties:
1. product rule :- $$\log \left( {mn} \right) = \log m + \log n$$
2. Quotient rule :- $$\log \left( {\dfrac{m}{n}} \right) = \log m - \log n$$
3. Power rule :- $$\log \left( {{m^n}} \right) = n.\log m$$
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