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Chapter 1: Problem 17

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

Short Answer

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

Step by step solution

01

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: \[ \frac{1}{\log _{5} 3} \cdot \log _{25} 27 \]
02

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: \[ \frac{1}{\log _{5} 3} \cdot \left(\frac{\log_{5}{27}}{\log_{5}{25}}\right) \]
03

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: \[ \frac{1}{\log _{5} 3} \cdot \frac{3\log_{5}3}{2\log_{5}5} \]
04

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 \(\frac{3}{2}\).

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