Problem 17 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \log _{3} 5 \cdot \log _{25} 27 $$

Step-by-Step Solution

Answer

The simplified expression for \(\log _{3} 5 \cdot \log _{25} 27\) is \(\frac{3}{2}\).

Step 1: Utilizing the Rule \(\log_{a}b = \frac{1}{\log_{b}a}\)

Rewrite the expression \(\log _{3} 5\) as \(\frac{1}{\log _{5} 3}\) so, the expression becomes: \[ \frac{1}{\log _{5} 3} \cdot \log _{25} 27 \]

Step 2: Applying the Change of Base Formula

Now, apply the change of base formula to the second logarithmic part of the expression. We want to match the base of the second expression to the number 5 from the first expression. \(\log_{25}27\) becomes \(\frac{\log_{5}{27}}{\log_{5}{25}}\). Now, the expression becomes: \[ \frac{1}{\log _{5} 3} \cdot \left(\frac{\log_{5}{27}}{\log_{5}{25}}\right) \]

Step 3: Simplification

Simplify \(\log_{5} 27\) and \(\log_{5} 25\). Since \(27 = 3^3\) and \(25 = 5^2\), \(\log_{5} 27\) and \(\log_{5} 25\), will simplify to \(3\log_{5}3\) and \(2\log_{5}5\) respectively. Now, the expression becomes: \[ \frac{1}{\log _{5} 3} \cdot \frac{3\log_{5}3}{2\log_{5}5} \]

Step 4: Cancellation

Now, cancel out common terms. The \(\log _{5} 3\) in the numerator and the denominator cancel out and likewise for \(\log _{5} 5\). The expression simplifies to: \[ \frac{3}{2} \] So, the expression \(\log _{3} 5 \cdot \log _{25} 27\) simplifies to \(\frac{3}{2}\).