Prove that `log_(2)17.log_(1//5)2.log_(3).(1)/(5)gt2`

Answer

To prove that \( \log_{2} 17 \cdot \log_{\frac{1}{5}} 2 \cdot \log_{3} \frac{1}{5} > 2 \), we will follow these steps:

### Step 1: Rewrite the logarithms
We can use the change of base formula for logarithms, which states that \( \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \). Here, we will use base 10 for convenience.

\[ \log_{2} 17 = \frac{\log 17}{\log 2} \]
\[ \log_{\frac{1}{5}} 2 = \log_{5^{-1}} 2 = -\log_{5} 2 = -\frac{\log 2}{\log 5} \]
\[ \log_{3} \frac{1}{5} = \log_{3} 5^{-1} = -\log_{3} 5 = -\frac{\log 5}{\log 3} \]

### Step 2: Substitute the rewritten logarithms into the inequality
Now substitute these expressions back into the original inequality:

\[ \log_{2} 17 \cdot \log_{\frac{1}{5}} 2 \cdot \log_{3} \frac{1}{5} = \left( \frac{\log 17}{\log 2} \right) \cdot \left( -\frac{\log 2}{\log 5} \right) \cdot \left( -\frac{\log 5}{\log 3} \right) \]

### Step 3: Simplify the expression
The negative signs will cancel out:

\[ = \frac{\log 17}{\log 2} \cdot \frac{\log 2}{\log 5} \cdot \frac{\log 5}{\log 3} \]

Notice that \( \log 2 \) and \( \log 5 \) cancels out:

\[ = \frac{\log 17}{\log 3} \]

### Step 4: Establish the inequality
We need to show that:

\[ \frac{\log 17}{\log 3} > 2 \]

This can be rewritten as:

\[ \log 17 > 2 \log 3 \]

Using the properties of logarithms, we can express \( 2 \log 3 \) as \( \log 3^2 \):

\[ \log 17 > \log 9 \]

### Step 5: Compare the values
Since \( 17 > 9 \), we can conclude that:

\[ \log 17 > \log 9 \]

Thus, we have proved that:

\[ \log_{2} 17 \cdot \log_{\frac{1}{5}} 2 \cdot \log_{3} \frac{1}{5} > 2 \]

### Conclusion
The original statement is proven true.

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