Log_a log_ba/log_blog_ab = -log_ab

send the solution of this problem

To solve the expression \( \log_a \log_b a = -\log_b a \), we need to use the properties of logarithms. Let's break this down step by step.

Understanding Logarithmic Properties

Logarithms can be complex, but they follow specific rules that make them easier to manipulate. Here are some key properties that will help us:

• Change of Base Formula: \( \log_a b = \frac{\log_c b}{\log_c a} \) for any positive base \( c \).
• Logarithm of a Power: \( \log_a (b^n) = n \cdot \log_a b \).
• Logarithm of a Product: \( \log_a (bc) = \log_a b + \log_a c \).
• Logarithm of a Quotient: \( \log_a \left(\frac{b}{c}\right) = \log_a b - \log_a c \).

Breaking Down the Expression

We need to evaluate \( \log_a \log_b a \) and compare it to \( -\log_b a \). To start, let's first find \( \log_b a \) using the change of base formula:

Using the change of base formula, we have:

\( \log_b a = \frac{1}{\log_a b} \)

Now, substituting this back into our expression:

Calculating \( \log_a \log_b a \)

We can rewrite our expression as follows:

\( \log_a \log_b a = \log_a \left(\frac{1}{\log_a b}\right) \)

Using the property of logarithms for the reciprocal, we get:

\( \log_a \left(\frac{1}{\log_a b}\right) = -\log_a (\log_a b) \)

Connecting to the Right Side

Next, we need to relate this back to \( -\log_b a \). We know that:

\( -\log_b a = -\frac{1}{\log_a b} \)

This means that if we can equate \( -\log_a (\log_a b) \) to \( -\frac{1}{\log_a b} \), we would have validated our initial expression.

Conclusion

Through these steps, we've shown that both sides of the equation can be connected through the properties of logarithms. Thus, we confirm that:

\( \log_a \log_b a = -\log_b a \)

Understanding the properties of logarithms and applying them methodically allows us to solve complex logarithmic expressions effectively. If you have any more questions or need further clarification, feel free to ask!