Problem 41 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left(\log _{\sqrt{5}} 125 \div \log _{5}^{2} 25\right) \cdot\left(\log _{\frac{1}{5}} \sqrt{5} \div \log _{0.2} \sqrt[3]{25}\right) $$

Step-by-Step Solution

Answer

The simplified expression is equal to \(6\).

Step 1: Simplify the first fraction given by \(\frac{\log _{\sqrt{5}} 125}{\log _{5}^{2} 25}\)

We can rewrite this fraction using the change of base formula for logarithms. Let's use base 10 logarithms to simplify, but you can use any base as long as it's consistent for the two logarithms within the fraction: \[\frac{\log _{\sqrt{5}} 125}{\log _{5}^{2} 25} = \frac{\frac{\log_{10} 125}{\log_{10} \sqrt{5}}}{\frac{\log_{10} 25}{\log_{10} 5^2}}\]

Step 2: Simplify the expression further

Now simplify the inner fractions: \[\frac{\log_{10} 125}{\log_{10} \sqrt{5}} \cdot \frac{\log_{10} 5^2}{\log_{10} 25}\]

Step 3: Calculate the logarithms

We'll now use the properties of logarithms to compute the values: - \(\log_{10} 125 = 3 \log_{10} 5\) - \(\log_{10} \sqrt{5} = \frac{1}{2} \log_{10} 5\) - \(\log_{10} 5^2 = 2 \log_{10} 5\) - \(\log_{10} 25 = 2 \log_{10} 5\) Now replace those values in our expression: \[\frac{3 \log_{10} 5}{\frac{1}{2} \log_{10} 5} \cdot \frac{2 \log_{10} 5}{2 \log_{10} 5}\]

Step 4: Simplify and cancel the factors

Cancel out the common factors: \[\frac{6}{1} \cdot \frac{1}{1} = 6\] The simplified expression is equal to \(6\).

Key Concepts

Change of Base Formula

The change of base formula is a crucial tool in simplifying logarithmic expressions. It allows us to rewrite a logarithm in terms of logs of a different base. This is particularly useful when we have logs with uncommon bases or if a problem requires consistent calculation. The formula is given as:

• \(rac{\log_a b}{\log_a c} = \frac{\log_k b}{\log_k c}\)

This equation lets you select a convenient base \(k\), like 10 or \(e\). This is applicable wherever maintaining consistent base is necessary. In our example, using base 10 made it easier to compute: converting \(\log_{\sqrt{5}} 125\) and \(\log_5^2 25\) into a manageable format through conversion to base 10.

Properties of Logarithms

Understanding logarithmic properties is essential in simplifying complicated expressions. Key properties include:

• Product Property: \(\log_b (mn) = \log_b m + \log_b n\)
• Quotient Property: \(\log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n\)
• Power Property: \(\log_b (m^n) = n \cdot \log_b m\)

These allow you to break down and simplify expressions efficiently. The transformation of \(\log_{10} 5^2\) and \(\log_{10} 25\) was possible by applying the power property, where the exponents were factored out, revealing the underlying base calculations.

Logarithmic Simplification

To simplify logarithmic expressions, especially those involving multiple operations, applying the right logarithmic properties is key. In the exercise, the fraction \(\frac{3 \log_{10} 5}{\frac{1}{2} \log_{10} 5}\) was simplified by canceling out \(\log_{10} 5\). This reduces complex expressions to simpler fractions. Using consistent bases and properties can convert challenging tasks into straightforward calculations. Simplification is about recognizing factor ability and cancelability, streamlining the problem without changing its original value.

Fraction Simplification

Fraction simplification involves reducing fractions by canceling common factors from the numerator and the denominator. When dealing with logarithmic fractions, ensure the terms are simplified beforehand. For instance:

• The expression \(\frac{6}{1} \cdot \frac{1}{1}\) results from canceling equal terms, effectively reducing the complexity down to a simple numerical answer of 6.

This is crucial for correctly interpreting logarithmic terms where equivalent logarithmic values are often hidden within complex structures. Effective fraction simplification turns complicated structures into straightforward numbers, highlighting the essence of the calculation.