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Question

Question: How do you solve\[\ln \left( x \right)-4\ln 3=\ln \left( \dfrac{5}{x} \right)\]?...

How do you solveln(x)4ln3=ln(5x)\ln \left( x \right)-4\ln 3=\ln \left( \dfrac{5}{x} \right)?

Explanation

Solution

In the given question, we have been asked to solveln(x)4ln3=ln(5x)\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 alnb=ln(ba)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 alnb=ln(ba)a\ln b=\ln \left( {{b}^{a}} \right).
● The quotient property of logarithm which states that lnalnb=ln(ab)\ln a-\ln b=\ln \left( \dfrac{a}{b} \right).

Complete step by step solution:
We have given that,
ln(x)4ln3=ln(5x)\ln \left( x \right)-4\ln 3=\ln \left( \dfrac{5}{x} \right)
Using the property of logarithm, i.e.
alnb=ln(ba)a\ln b=\ln \left( {{b}^{a}} \right)
Applying the property, we get
ln(x)ln(34)=ln(5x)\ln \left( x \right)-\ln \left( {{3}^{4}} \right)=\ln \left( \dfrac{5}{x} \right)
Using the quotient property of logarithm, i.e.
lnalnb=ln(ab)\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)
Applying the quotient property of log, we get
ln(x34)=ln(5x)\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
eln(x34)=eln(5x)\Rightarrow {{e}^{\ln \left( \dfrac{x}{{{3}^{4}}} \right)}}={{e}^{\ln \left( \dfrac{5}{x} \right)}}
As we know that, eln(x)=x{{e}^{\ln \left( x \right)}}=x
Thus,
x34=5x\Rightarrow \dfrac{x}{{{3}^{4}}}=\dfrac{5}{x}
Cross multiplication in the above equation, we get
x2=5×34\Rightarrow {{x}^{2}}=5\times {{3}^{4}}
Simplifying the above equation, we get
x=±5×34=±5×32\Rightarrow x=\pm \sqrt{5\times {{3}^{4}}}=\pm \sqrt{5}\times {{3}^{2}}
Therefore,
x=+5×32=95\Rightarrow x=+\sqrt{5}\times {{3}^{2}}=9\sqrt{5} and x=5×32=95x=-\sqrt{5}\times {{3}^{2}}=-9\sqrt{5}
As the log function cannot take negative values, so x = 95-9\sqrt{5} is not a solution.
Thus, the possible value of ‘x’ is 959\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’.