How do you solve\[\ln \left( x \right)-4\ln 3=\ln \left( \dfrac{5}{x} \right)\]?
Answer
621k+ views
Hint:In the given question, we have been asked to solve\[\ln \left( x \right)-4\ln 3=\ln \left( \dfrac{5}{x} \right)\]. In order to solve the question for the value of ‘x’, first we apply the property of logarithm which states that \[a\ln b=\ln \left( {{b}^{a}} \right)\]and later by applying the quotient property of logarithmic function we will simplify further. Then to eliminate the log function we need to convert the logarithmic equation into the exponential equation and solve the equation further in a way we solve the general equation.
Formula used:
● The property of logarithm, states that \[a\ln b=\ln \left( {{b}^{a}} \right)\].
● The quotient property of logarithm which states that \[\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)\].
Complete step by step solution:
We have given that,
\[\ln \left( x \right)-4\ln 3=\ln \left( \dfrac{5}{x} \right)\]
Using the property of logarithm, i.e.
\[a\ln b=\ln \left( {{b}^{a}} \right)\]
Applying the property, we get
\[\ln \left( x \right)-\ln \left( {{3}^{4}} \right)=\ln \left( \dfrac{5}{x} \right)\]
Using the quotient property of logarithm, i.e.
\[\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)\]
Applying the quotient property of log, we get
\[\Rightarrow \ln \left( \dfrac{x}{{{3}^{4}}} \right)=\ln \left( \dfrac{5}{x} \right)\]
To eliminate log function or to cancel out the log function, we raise ‘e’ to the power log.
Converting the logarithmic equation into exponential form, we get
\[\Rightarrow {{e}^{\ln \left( \dfrac{x}{{{3}^{4}}} \right)}}={{e}^{\ln \left( \dfrac{5}{x} \right)}}\]
As we know that, \[{{e}^{\ln \left( x \right)}}=x\]
Thus,
\[\Rightarrow \dfrac{x}{{{3}^{4}}}=\dfrac{5}{x}\]
Cross multiplication in the above equation, we get
\[\Rightarrow {{x}^{2}}=5\times {{3}^{4}}\]
Simplifying the above equation, we get
\[\Rightarrow x=\pm \sqrt{5\times {{3}^{4}}}=\pm \sqrt{5}\times {{3}^{2}}\]
Therefore,
\[\Rightarrow x=+\sqrt{5}\times {{3}^{2}}=9\sqrt{5}\] and \[x=-\sqrt{5}\times {{3}^{2}}=-9\sqrt{5}\]
As the log function cannot take negative values, so x = \[-9\sqrt{5}\] is not a solution.
Thus, the possible value of ‘x’ is \[9\sqrt{5}\] .
It is the required solution.
Note: In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general linear equations. Students should always remember that natural log and the exponential functions are the inverse of each other, which means that if we raise the exponential function by the natural log of x, then only we would be able to find the value of ‘x’.
Formula used:
● The property of logarithm, states that \[a\ln b=\ln \left( {{b}^{a}} \right)\].
● The quotient property of logarithm which states that \[\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)\].
Complete step by step solution:
We have given that,
\[\ln \left( x \right)-4\ln 3=\ln \left( \dfrac{5}{x} \right)\]
Using the property of logarithm, i.e.
\[a\ln b=\ln \left( {{b}^{a}} \right)\]
Applying the property, we get
\[\ln \left( x \right)-\ln \left( {{3}^{4}} \right)=\ln \left( \dfrac{5}{x} \right)\]
Using the quotient property of logarithm, i.e.
\[\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)\]
Applying the quotient property of log, we get
\[\Rightarrow \ln \left( \dfrac{x}{{{3}^{4}}} \right)=\ln \left( \dfrac{5}{x} \right)\]
To eliminate log function or to cancel out the log function, we raise ‘e’ to the power log.
Converting the logarithmic equation into exponential form, we get
\[\Rightarrow {{e}^{\ln \left( \dfrac{x}{{{3}^{4}}} \right)}}={{e}^{\ln \left( \dfrac{5}{x} \right)}}\]
As we know that, \[{{e}^{\ln \left( x \right)}}=x\]
Thus,
\[\Rightarrow \dfrac{x}{{{3}^{4}}}=\dfrac{5}{x}\]
Cross multiplication in the above equation, we get
\[\Rightarrow {{x}^{2}}=5\times {{3}^{4}}\]
Simplifying the above equation, we get
\[\Rightarrow x=\pm \sqrt{5\times {{3}^{4}}}=\pm \sqrt{5}\times {{3}^{2}}\]
Therefore,
\[\Rightarrow x=+\sqrt{5}\times {{3}^{2}}=9\sqrt{5}\] and \[x=-\sqrt{5}\times {{3}^{2}}=-9\sqrt{5}\]
As the log function cannot take negative values, so x = \[-9\sqrt{5}\] is not a solution.
Thus, the possible value of ‘x’ is \[9\sqrt{5}\] .
It is the required solution.
Note: In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general linear equations. Students should always remember that natural log and the exponential functions are the inverse of each other, which means that if we raise the exponential function by the natural log of x, then only we would be able to find the value of ‘x’.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

