
Is \[\left({ln\ x} \right)^{2}\] equivalent to \[ln^{2}\left( x \right)\] ?
Answer
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Hint: In this question, we need to find whether \[\left({ln\ x} \right)^{2}\] is equivalent to \[ln^{2}\left( x \right)\] . ln is nothing but it is a natural log that is log with base \[e\]. \[log{_e}= \ ln\] (natural log). Here \[e\] is the exponential function. Mathematically, ln is known as natural logarithm and it is also known as logarithm of base \[e\] . Mathematically it is represented as \[{ln\ x}\] or \[\log{_e} x\] . Logarithm is nothing but a power to which numbers must be raised to get some other values. Here we need to find whether \[\left({ln\ x} \right)^{2}\] is equivalent to \[ln^{2}\left( x \right)\] .
Logarithmic property used :
\[{log\ }\left( m^{n} \right) = \ n\ log\ m\]
Complete step-by-step solution:
We need to find whether \[\left( {ln\ x} \right)^{2}\] is equivalent to \[ln^{2}\left( x \right)\]
In order to find whether \[\left( {ln\ x} \right)^{2}\] and \[ln^{2}\left( x \right)\] are equivalent, first we can assume that \[ln^{2}\left( x \right) = \left( {ln\ x} \right)^{2}\]
Let us assume that
\[ln^{2}\left( x \right) = \left( {ln\ x} \right)^{2}\]
From the property of logarithm,
\[{log\ }\left( m^{n} \right) = \ n\ log\ m\]
We get,
\[\Rightarrow (\ln{\ x)}^{2} = 2\ln x\]
\[(\ln{\ x)}^{2}\] is \[ln\ x \times ln\ x\] Since \[ln\ x\ \neq 0\]
\[\Rightarrow ln\ x \times ln\ x = 2\ln x\]
On dividing both sides by \[ln\ x\] ,
We get,
\[{ln\ x} = 2\]
On taking exponential on both sides we get,
\[x = e^{2}\]
Hence we can conclude that \[ln^{2}\left( x \right) = \left( {ln\ x} \right)^{2}\] is true for \[x = e^{2}\]
Final answer :
\[\left({ln\ x} \right)^{2}\] is equivalent to \[ln^{2}\left( x \right)\]
Note: We need to know that the logarithmic function to the base \[e\] is known as the natural logarithmic function and it is denoted by \[\log{_e}\]. We should not be confused \[{ln\ }x^{2}\] with \[2ln\ x\] where both are the same. The inverse of logarithm is known as exponential. Exponential function is nothing but it is a mathematical function which is in the form of \[f\ (x)\ = \ a^{x}\], where \[x\] is a variable and \[a\] is a constant . The most commonly used exponential base is \[e\] which is approximately equal to \[2.71828\] .
Few properties of logarithm are
1.\[log\ mn\ = \ log\ m\ + \ log\ n\]
2.\[{log\ }\left( \dfrac{m}{n} \right) = \ log\ m\ \ log\ n\]
3.\[{log\ }\left( m^{n} \right) = \ n\ log\ m\]
Logarithmic property used :
\[{log\ }\left( m^{n} \right) = \ n\ log\ m\]
Complete step-by-step solution:
We need to find whether \[\left( {ln\ x} \right)^{2}\] is equivalent to \[ln^{2}\left( x \right)\]
In order to find whether \[\left( {ln\ x} \right)^{2}\] and \[ln^{2}\left( x \right)\] are equivalent, first we can assume that \[ln^{2}\left( x \right) = \left( {ln\ x} \right)^{2}\]
Let us assume that
\[ln^{2}\left( x \right) = \left( {ln\ x} \right)^{2}\]
From the property of logarithm,
\[{log\ }\left( m^{n} \right) = \ n\ log\ m\]
We get,
\[\Rightarrow (\ln{\ x)}^{2} = 2\ln x\]
\[(\ln{\ x)}^{2}\] is \[ln\ x \times ln\ x\] Since \[ln\ x\ \neq 0\]
\[\Rightarrow ln\ x \times ln\ x = 2\ln x\]
On dividing both sides by \[ln\ x\] ,
We get,
\[{ln\ x} = 2\]
On taking exponential on both sides we get,
\[x = e^{2}\]
Hence we can conclude that \[ln^{2}\left( x \right) = \left( {ln\ x} \right)^{2}\] is true for \[x = e^{2}\]
Final answer :
\[\left({ln\ x} \right)^{2}\] is equivalent to \[ln^{2}\left( x \right)\]
Note: We need to know that the logarithmic function to the base \[e\] is known as the natural logarithmic function and it is denoted by \[\log{_e}\]. We should not be confused \[{ln\ }x^{2}\] with \[2ln\ x\] where both are the same. The inverse of logarithm is known as exponential. Exponential function is nothing but it is a mathematical function which is in the form of \[f\ (x)\ = \ a^{x}\], where \[x\] is a variable and \[a\] is a constant . The most commonly used exponential base is \[e\] which is approximately equal to \[2.71828\] .
Few properties of logarithm are
1.\[log\ mn\ = \ log\ m\ + \ log\ n\]
2.\[{log\ }\left( \dfrac{m}{n} \right) = \ log\ m\ \ log\ n\]
3.\[{log\ }\left( m^{n} \right) = \ n\ log\ m\]
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