Question

The value of \(\frac{1}{\log_3{60}} + \frac{1}{\log_4{60}} + \frac{1}{\log_5{60}}\) is

• A. 0
• B. 1
• C. 5
• D. 60

Answer 1

The Correct Option is B

We need to evaluate the expression:

\[ \frac{1}{\log_3{60}} + \frac{1}{\log_4{60}} + \frac{1}{\log_5{60}} \] 

Step 1: Change of Base Formula

Using the logarithm identity \( \log_a b = \frac{\log b}{\log a} \), we rewrite each term:

\[ \frac{1}{\log_3{60}} = \frac{\log 3}{\log 60}, \quad \frac{1}{\log_4{60}} = \frac{\log 4}{\log 60}, \quad \frac{1}{\log_5{60}} = \frac{\log 5}{\log 60} \]

Step 2: Summation

Adding these terms:

\[ \frac{\log 3}{\log 60} + \frac{\log 4}{\log 60} + \frac{\log 5}{\log 60} \]

Factor out \( \frac{1}{\log 60} \):

\[ \frac{\log 3 + \log 4 + \log 5}{\log 60} \]

Step 3: Logarithm Properties

Using \( \log a + \log b = \log (a \times b) \):

\[ \log 3 + \log 4 + \log 5 = \log (3 \times 4 \times 5) = \log 60 \]

Thus, we get:

\[ \frac{\log 60}{\log 60} = 1 \]

Final Answer: 1

Answer 2

To find the value of \( \frac{1}{\log_3{60}} + \frac{1}{\log_4{60}} + \frac{1}{\log_5{60}} \), we start by using the change of base formula for logarithms:

\[ \frac{1}{\log_b{a}} = \log_a{b} \]

Applying this to each term, we have:

\[ \frac{1}{\log_3{60}} = \log_{60}{3} \]

\[ \frac{1}{\log_4{60}} = \log_{60}{4} \]

\[ \frac{1}{\log_5{60}} = \log_{60}{5} \]

Thus, combining these results, we get:

\[ \log_{60}{3} + \log_{60}{4} + \log_{60}{5} \]

Using the property of logarithms that states: \(\log_a{b} + \log_a{c} = \log_a{(b \cdot c)}\), we can combine these terms:

\[ \log_{60}{(3 \cdot 4 \cdot 5)} \]

Calculating the product, we have \(3 \cdot 4 \cdot 5 = 60\).

Thus, the expression simplifies to:

\[ \log_{60}{60} \]

Since \(\log_s{s} = 1\) for any \(s > 0\), we find that:

\[ \log_{60}{60} = 1 \]

Hence, the value of the given expression is \(1\).