If a =`log_3(5)` and b =`log_17( 25)`, which one of the following is correct?

Options

A. `altb`
B. a=b
C. `agtb`
D. None of these

Answer

To solve the problem, we need to compare the values of \( a \) and \( b \) given by:

\[ a = \log_3(5) \quad \text{and} \quad b = \log_{17}(25) \]

### Step 1: Rewrite \( a \) and \( b \) using the change of base formula

Using the change of base formula, we can express both logarithms in terms of natural logarithms (or common logarithms):

\[ a = \log_3(5) = \frac{\log(5)}{\log(3)} \]

\[ b = \log_{17}(25) = \frac{\log(25)}{\log(17)} \]

### Step 2: Simplify \( b \)

Since \( 25 = 5^2 \), we can simplify \( b \):

\[ b = \frac{\log(5^2)}{\log(17)} = \frac{2 \log(5)}{\log(17)} \]

### Step 3: Compare \( a \) and \( b \)

Now we have:

\[ a = \frac{\log(5)}{\log(3)} \quad \text{and} \quad b = \frac{2 \log(5)}{\log(17)} \]

To compare \( a \) and \( b \), we can analyze the fractions. We can express \( b \) in terms of \( a \):

\[ b = \frac{2 \log(5)}{\log(17)} = 2 \cdot \frac{\log(5)}{\log(17)} \]

### Step 4: Find a common denominator

To compare \( a \) and \( b \), we can find a common denominator:

\[ a = \frac{\log(5)}{\log(3)} \quad \text{and} \quad b = \frac{2 \log(5)}{\log(17)} \]

We can cross-multiply to compare:

\[ a \cdot \log(17) = \log(5) \cdot \log(17) \]
\[ b \cdot \log(3) = 2 \log(5) \cdot \log(3) \]

### Step 5: Set up the inequality

Now we set up the inequality:

\[ \log(5) \cdot \log(17) \quad \text{vs} \quad 2 \log(5) \cdot \log(3) \]

Assuming \( \log(5) > 0 \) (which it is), we can divide both sides by \( \log(5) \):

\[ \log(17) \quad \text{vs} \quad 2 \log(3) \]

### Step 6: Compare \( \log(17) \) and \( 2 \log(3) \)

We know that \( 2 \log(3) = \log(3^2) = \log(9) \). Therefore, we need to compare \( \log(17) \) and \( \log(9) \).

Since \( 17 > 9 \), we have:

\[ \log(17) > \log(9) \]

### Conclusion

Since \( \log(17) > 2 \log(3) \), we conclude that:

\[ b < a \]

Thus, the correct answer is:

\[ \text{Option 3: } b < a \]