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Question: How do you solve the logarithmic equation \({{\log }_{3}}x-2{{\log }_{x}}3=1\) ?...

How do you solve the logarithmic equation log⁡3x−2log⁡x3=1{{\log }_{3}}x-2{{\log }_{x}}3=1 ?

Explanation

Solution

Problems of this type can be solved by rewriting the logarithmic terms by using the change of base formula of logarithms. We elaborate the terms as a quotient of logarithmic terms having a common positive base. Thus, the equation becomes a quadratic equation which is further solved by factorization method.

Complete step-by-step solution:
The equation we have is
log⁡3x−2log⁡x3=1{{\log }_{3}}x-2{{\log }_{x}}3=1
As both the logarithmic terms have different bases, we use the change of base formula of logarithm.
According to the Change of base formula of logarithm
log⁡ba=log⁡alog⁡b{{\log }_{b}}a=\dfrac{\log a}{\log b}
Therefore, we rewrite the term log⁡3x{{\log }_{3}}x as log⁡xlog⁡3\dfrac{\log x}{\log 3} and the term log⁡x3{{\log }_{x}}3 as log⁡3log⁡x\dfrac{\log 3}{\log x}
Substituting the above terms in the given equation we get
⇒log⁡xlog⁡3−2log⁡3log⁡x=1\Rightarrow \dfrac{\log x}{\log 3}-2\dfrac{\log 3}{\log x}=1
We assume that log⁡xlog⁡3=y\dfrac{\log x}{\log 3}=y
Substituting the above expression with yy in the above equation we get
⇒y−2y=1\Rightarrow y-\dfrac{2}{y}=1
Multiplying the above equation with yy to both the sides we get
⇒y2−2=y\Rightarrow {{y}^{2}}-2=y
Further simplifying the above equation, we get
⇒y2−y−2=0\Rightarrow {{y}^{2}}-y-2=0
The above equation has become a quadratic equation. So, we can rewrite the coefficient of yy as
(2−1)=1\left( 2-1 \right)=1 , as 22 and 11 are the multipliers of 22
Hence, the equation can be written as
⇒y2−(2−1)y−2=0\Rightarrow {{y}^{2}}-\left( 2-1 \right)y-2=0
We now apply the distribution on the middle term of the expression and get
⇒y2−2y+y−2=0\Rightarrow {{y}^{2}}-2y+y-2=0
We can now take yy common from y2−2y{{y}^{2}}-2y and 11 common from the rest terms as
⇒y(y−2)+1(y−2)=0\Rightarrow y\left( y-2 \right)+1\left( y-2 \right)=0
Again, we can see that the term (y−2)\left( y-2 \right) can be taken as common from the entire expression as
⇒(y−2)(y+1)=0\Rightarrow \left( y-2 \right)\left( y+1 \right)=0
Hence,
y−2=0y-2=0
⇒y=2\Rightarrow y=2
Substituting the expression of yy we get
⇒log⁡xlog⁡3=2\Rightarrow \dfrac{\log x}{\log 3}=2
⇒log⁡x=2log⁡3\Rightarrow \log x=2\log 3
⇒log⁡x=log⁡9\Rightarrow \log x=\log 9
⇒x=9\Rightarrow x=9
Again, y+1=0y+1=0
⇒y=−1\Rightarrow y=-1
Substituting the expression of yy we get
⇒log⁡xlog⁡3=−1\Rightarrow \dfrac{\log x}{\log 3}=-1
⇒log⁡x=−log⁡3\Rightarrow \log x=-\log 3
⇒log⁡x=log⁡13\Rightarrow \log x=\log \dfrac{1}{3}
⇒x=13\Rightarrow x=\dfrac{1}{3}
Therefore, we conclude that the solutions of the given equation are x=9,13x=9,\dfrac{1}{3}.

Note: While changing the base of the logarithmic terms we must be very careful so that mistakes are avoided. Also, the quadratic equation we got can be solved by other methods, such as forming a square method or by directly using Sridhar Acharya's formula.