Problem 8 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Determine \(b\) satisfying (i) \(\log _{\sqrt{8}} b=3 \frac{1}{3}\) (ii) \(\log _{\sqrt{8}} b=3^{\frac{1}{3}}\) (iii) \(\log _{a} 2 \cdot \log _{b} 625=\log _{10} 16 \cdot \log _{a} 10\) (iv) \(\log _{3} \log _{2} \log _{\sqrt{5}}(625)=b\)

Step-by-Step Solution

Answer

The values of \(b\) that satisfies the different logarithmic equations are \(b = 8\) , \(b = \sqrt{8}\), for any positive number, and \(b = 0.43\) respectively.

Step 1: Solve the logarithmic equation (i)

The logarithmic equation can be rewritten as an exponential equation. In other words, \( \sqrt{8}^{3 \frac{1}{3}} = b \) which results in \( b = 8\).

Step 2: Solve the logarithmic equation (ii)

The logarithmic equation can be rewritten as an exponential equation. In other words, \( \sqrt{8}^{3^{\frac{1}{3}}} = b \) which results in \( b = \sqrt{8} \).

Step 3: Solve the logarithmic equation (iii)

As logarithmic equations are equivalent under the change of base formula, it is possible to set \( a = 2 \) and \( b = 625 \). In other words, \( \log _{2} 2 \cdot \log _{625} 625 = \log _{10} 16 \cdot \log _{2} 10 \) This simplifies to \( 1 \cdot 1 = \log _{10} 2^4 \cdot \frac{1}{\log _{10} 2} \).And after performing the calculations, it simplifies to \( 1 = 1 \). This means that the logarithmic equation holds true for any positive value of \( b \).

Step 4: Solve the logarithmic equation (iv)

The nested logarithms in the equation can be resolved in stages. This gives \( b = \log _{3} \log _{2} 5 \) and \( b = \log _{3} 2.32 = 0.43\).

Key Concepts

Exponential Equations

Exponential equations involve expressions where variables appear as exponents. These types of equations are crucial because they provide a foundation for understanding growth and decay in various fields, like physics, finance, and biology. In the context of our problem, converting a logarithmic equation into an exponential form simplifies solving the equation.
Let's consider the equation from the original exercise:

• For (i): \[\log _{\sqrt{8}} b = 3\frac{1}{3}\] This becomes: \[\sqrt{8}^{3\frac{1}{3}} = b\]Here, the logarithm is converted to an exponential form, making it easier to solve for \(b\) directly. The solution process involves raising the base \(\sqrt{8}\) to the power \(3\frac{1}{3}\), resulting in \(b = 8\).

Each logarithmic equation can typically be converted to an exponential equation for easy resolution. This is especially useful in complex equations where directly solving a logarithm might be difficult. Understanding the connection between the two forms is fundamental to mastering logarithmic and exponential problems.

Change of Base Formula

The change of base formula is a helpful tool for solving logarithmic equations by expressing a logarithm in terms of logs of different bases. It offers flexibility and is especially useful when calculators only support logarithms of a particular base, like base 10.
For equation (iii) in the exercise:

• The equation \[\log _{a} 2 \cdot \log _{b} 625 = \log _{10} 16 \cdot \log _{a} 10\]is resolved by selecting specific values for \(a\) and \(b\) that simplify the equation using the change of base formula.
• By letting \(a = 2\) and \(b = 625\), each part of the equation simplifies, validating that it holds true universally. This showcases the power of the change of base formula in proving or simplifying logarithmic equalities.

The change of base formula is expressed as:\[\log _{b} a = \frac{\log _{c} a}{\log _{c} b}\]where \(c\) is a new base, often chosen for convenience depending on the problem. Mastering this allows for greater versatility in solving logarithmic problems.

Nested Logarithms

Nested logarithms are logarithms within other logarithms. They often appear in problems requiring multiple steps to solve. Understanding how to "unwrap" these logarithms is critical in solving such complexities.
In equation (iv) from the exercise:

• \[ \log _{3} \log _{2} \log _{\sqrt{5}}(625) = b \]
Unwrapping this begins at the innermost logarithm:
• Calculate \(\log _{\sqrt{5}}(625)\) first. Next, find \(\log _{2}\) of that result,.Finally, calculate the outermost \(\log _{3}\) to get \(b\).

This equation simplifies step-by-step:

First: \(b = \log _{3} \log _{2} 5\)

Then: \(b = \log _{3} 2.32\).

Solving these nested parts gives \(b \approx 0.43\).

Nested logarithms require patience and practice. Start from the innermost part, and systematically work your way outwards to simplify the expression fully.

Regular Logarithms

Regular logarithms, or base-10 logarithms, are the most common logarithmic form. They're essential in various scientific fields, since they simplify calculations significantly when dealing with extremely large or small numbers.
In our exercise:

• Part (iii) particularly makes use of base-10 logic as seen in: \[\log _{10} 16 \cdot \log _{a} 10\]
Recognizing common logarithmic bases, especially base 10, makes handling equations more straightforward.
Important properties of regular logarithms include:
• \(\log_{10}(10) = 1\) - This stems from the fact that \(10^1 = 10\).
• The ability to easily compute exponents in forms like \(10^n\), reflecting how many times 10 is multiplied by itself.

This makes base-10 logarithms practical for manual calculations and easily understood even by rudimentary calculators.