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Question: Solve the logarithmic expression \[{{2}^{{{\log }_{3}}5}}-{{5}^{{{\log }_{3}}2}}\]....

Solve the logarithmic expression 2log⁡35−5log⁡32{{2}^{{{\log }_{3}}5}}-{{5}^{{{\log }_{3}}2}}.

Explanation

Solution

Hint: First assume 3 variables a, b, c. Now take the term alog⁡bc{{a}^{{{\log }_{b}}c}}. Now assume this term as ‘k’. Apply logarithm on both sides. Now you get a product of 2 logarithmic terms that are equal to the logarithm term. Now manipulate the term on the right-hand side to get the relation between the first assumed term and this manipulated term. Now substitute the given values in the equation to get the result we require in this question. Use the basic logarithmic properties in the process of manipulation. Simplify till you replace the positions of a, c in the first term.

Complete step-by-step solution-
Let us assume 3 variables which are positive integers: a, b, c.
Now take another variable k, which can be written as: k=alog⁡bck={{a}^{{{\log }_{b}}c}}.
Apply logarithm on both sides of the above equations, we get it as:
\Rightarrow $$$$\log k={{\log }_{b}}c\log a.
According to basic logarithmic properties, we know the formula:
\Rightarrow $$$${{\log }_{b}}c=\dfrac{\log c}{\log b}.
By substituting this into our first equation, we get it as:
\Rightarrow $$$$\log k=\dfrac{\log c}{\log b}\log a.
By using the property of a. b = b. a, we can use this and write:
\Rightarrow $$$$\log k=\dfrac{\log a }{\log b}\log c.
By using basic logarithmic properties, we know the formula of:
\Rightarrow $$$$\dfrac{\log c }{\log b}={{\log }_{b}}c
By substituting this in our equation, we get it as:
⇒log⁡k=log⁡balog⁡c\Rightarrow \log k={{\log }_{b}}a\log c
We can write the product of a term and logarithm as follows:
⇒log⁡k=log⁡clog⁡ba\Rightarrow \log k=\log {{c}^{{{\log }_{b}}a}}
By substituting the original assumption of variable k, we get:
⇒log⁡alog⁡bc=log⁡clog⁡ba\Rightarrow \log {{a}^{{{\log }_{b}}c}}=\log {{c}^{{{\log }_{b}}a}}
Now removing logarithm operator on both sides of the equation, we get:
⇒alog⁡bc=clog⁡ba\Rightarrow {{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}
By substituting a = 5, b = 3, c =2, we get the equations:
⇒5log⁡32=2log⁡35\Rightarrow {{5}^{{{\log }_{3}}2}}={{2}^{{{\log }_{3}}5}}
By subtracting the term 5log⁡32{{5}^{{{\log }_{3}}2}} on both sides, we get it as:
⇒2log⁡35−5log⁡32=0\Rightarrow {{2}^{{{\log }_{3}}5}}-{{5}^{{{\log }_{3}}2}}=0
So, the required result in the question is 0.
Therefore, we solved the required expression.

Note: Be careful with assumptions you take, if you misplace one of the variables you get the wrong result. While substituting the formula of the division of 2 terms, be careful. Solve the whole assumption till we get the power required term. We do this, all because in the question we have 2, 5 in opposite positions in both terms.