Prove that `log_7`11 is greater than ` log_(8)5` .

Answer

To prove that \( \log_7{11} > \log_8{5} \), we will use the change of base formula for logarithms. The change of base formula states that:

\[ \log_a{b} = \frac{\log_c{b}}{\log_c{a}} \]

for any positive base \( c \).

### Step 1: Apply the change of base formula

We will convert both logarithms to base 10 (or natural logarithm base \( e \)).

\[ \log_7{11} = \frac{\log_{10}{11}}{\log_{10}{7}} \]
\[ \log_8{5} = \frac{\log_{10}{5}}{\log_{10}{8}} \]

### Step 2: Set up the inequality

We need to show that:

\[ \frac{\log_{10}{11}}{\log_{10}{7}} > \frac{\log_{10}{5}}{\log_{10}{8}} \]

### Step 3: Cross-multiply to eliminate the fractions

Cross-multiplying gives us:

\[ \log_{10}{11} \cdot \log_{10}{8} > \log_{10}{5} \cdot \log_{10}{7} \]

### Step 4: Evaluate the logarithms

Now we will evaluate the logarithms using a calculator or logarithm tables:

- \( \log_{10}{11} \approx 1.0414 \)
- \( \log_{10}{8} \approx 0.9030 \)
- \( \log_{10}{5} \approx 0.6990 \)
- \( \log_{10}{7} \approx 0.8451 \)

### Step 5: Substitute the values into the inequality

Substituting these values into our inequality:

\[ 1.0414 \cdot 0.9030 > 0.6990 \cdot 0.8451 \]

Calculating both sides:

Left-hand side:
\[ 1.0414 \cdot 0.9030 \approx 0.9405 \]

Right-hand side:
\[ 0.6990 \cdot 0.8451 \approx 0.5890 \]

### Step 6: Compare the results

Since \( 0.9405 > 0.5890 \), we conclude that:

\[ \log_{10}{11} \cdot \log_{10}{8} > \log_{10}{5} \cdot \log_{10}{7} \]

### Conclusion

Thus, we have shown that:

\[ \log_7{11} > \log_8{5} \]

This completes the proof.

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