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Loga logba/logblogab = -logab
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Neeraj , 6 Years ago
Grade 12th pass
anser 1 Answers
Saurabh Koranglekar

Last Activity: 6 Years ago

To solve the expression logalogba=logba, 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: logab=logcblogca for any positive base c.
  • Logarithm of a Power: loga(bn)=nlogab.
  • Logarithm of a Product: loga(bc)=logab+logac.
  • Logarithm of a Quotient: loga(bc)=logablogac.

Breaking Down the Expression

We need to evaluate logalogba and compare it to logba. To start, let's first find logba using the change of base formula:

Using the change of base formula, we have:

logba=1logab

Now, substituting this back into our expression:

Calculating logalogba

We can rewrite our expression as follows:

logalogba=loga(1logab)

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

loga(1logab)=loga(logab)

Connecting to the Right Side

Next, we need to relate this back to logba. We know that:

logba=1logab

This means that if we can equate loga(logab) to 1logab, 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:

logalogba=logba

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!

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