If `x=log_5(1000)` and `y=log_7(2058)`,then

Options

A. `x gt y`
B. `x lt y`
C. `x=y`
D. none of these

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) given by:

1. \( x = \log_5(1000) \)
2. \( y = \log_7(2058) \)

### Step-by-Step Solution:

**Step 1: Calculate \( x = \log_5(1000) \)**

We can use the change of base formula for logarithms:
\[ \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \]
where \( k \) can be any positive number. Here, we will use base 10.

Thus, we can write:
\[ x = \log_5(1000) = \frac{\log_{10}(1000)}{\log_{10}(5)} \]

**Step 2: Simplify \( \log_{10}(1000) \)**

Since \( 1000 = 10^3 \), we have:
\[ \log_{10}(1000) = 3 \]

**Step 3: Find \( \log_{10}(5) \)**

Using a calculator or logarithm table, we find:
\[ \log_{10}(5) \approx 0.699 \]

**Step 4: Substitute back to find \( x \)**

Now substituting back, we get:
\[ x = \frac{3}{\log_{10}(5)} = \frac{3}{0.699} \approx 4.29 \]

**Step 5: Calculate \( y = \log_7(2058) \)**

Using the change of base formula again:
\[ y = \log_7(2058) = \frac{\log_{10}(2058)}{\log_{10}(7)} \]

**Step 6: Find \( \log_{10}(2058) \)**

Using a calculator, we find:
\[ \log_{10}(2058) \approx 3.313 \]

**Step 7: Find \( \log_{10}(7) \)**

Using a calculator or logarithm table, we find:
\[ \log_{10}(7) \approx 0.845 \]

**Step 8: Substitute back to find \( y \)**

Now substituting back, we get:
\[ y = \frac{3.313}{0.845} \approx 3.92 \]

**Step 9: Compare \( x \) and \( y \)**

Now we have:
- \( x \approx 4.29 \)
- \( y \approx 3.92 \)

Since \( x > y \), we conclude that:
\[ x > y \]

### Final Answer:
Thus, \( x \) is greater than \( y \).

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