Problem 30 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(x=\log _{a} b c, y=\log _{b} c a\) and \(z=\log _{c} a b\), then find the value of \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\)

Step-by-Step Solution

Answer

The value of \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\) is 1.

Step 1: Write the equations in exponential form

Using the definition of the logarithm, the given expressions can be rewritten as \(a^{x} = bc\), \(b^{y} = ca\), and \(c^{z} = ab\). This will simplify the expressions and make the problem easier to handle.

Step 2: Combine the equations

Multiplying all three equations together, we have \(a^{x} * b^{y} * c^{z} = bc * ca * ab = abcabc = (abc)^2\). This simplifies to \(a^{x}b^{y}c^{z} = (abc)^2\)

Step 3: Take log on both sides

To simplify further, take the log on both sides: \(\log(a^{x}b^{y}c^{z}) = \log((abc)^2)\). Using log properties, can be written as \(x\log(a) + y\log(b) + z\log(c) = 2(\log(a) + \log(b) + \log(c))\). Since \(\log(a) = z\), \(\log(b) = x\), and \(\log(c) = y\), this simplifies to \(2(x + y + z) = 2(x + y + z)\).

Step 4: Determine the formula value

Next, substitute \(x\), \(y\), and \(z\) into the equation \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\) and simplify. Since \(x + y + z = x + y + z\), the equation simplifies to 1

Key Concepts

Exponential Form in Logarithms

Understanding the exponential form of logarithms is crucial when dealing with logarithmic expressions in trigonometry. In essence, logarithms are the inverse operations of exponentiation. This means that if you have an equation like \( a^x = b \), you can express this in logarithmic form as \( x = \log_a b \).

Let's apply this to our exercise where you have \( x = \log_a bc \). By definition, this means that \( a^x = bc \). Seeing the logarithmic expression in its exponential form can clarify the relationship between the base \( a \), the exponent \( x \), and the result \( bc \).

The entire process of converting logarithms to their exponential form is a strategic move to make equations more workable. For instance, the step 1 of the provided solution uses this principle to set the stage for further manipulation of the given expressions. By recognizing this form, we gain a clear pathway to solving complex logarithmic expressions and this core understanding is imperative for tackling more advanced problems in mathematics.

Properties of Logarithms

Logarithms have unique properties that make them incredibly useful in simplifying expressions and solving equations. Three primary properties are vital to grasp:

• The product rule: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• The quotient rule: \( \log_b(\frac{M}{N}) = \log_b(M) - \log_b(N) \)
• The power rule: \( \log_b(M^k) = k \cdot \log_b(M) \)

During step 3 of our solution, these properties are utilized when taking the logarithm of both sides of the equation after combining the individual expressions. The property that allows us to write \( \log(a^xb^yc^z) \) as \( x\log(a) + y\log(b) + z\log(c) \) is an application of the product rule. This is a substantial simplification because it breaks apart the compounded logarithm into a sum of individual terms.

Being proficient with these properties not only helps in solving logarithmic equations but also in understanding their behavior and patterns, which is a significant part of higher-level mathematics.

Solving Logarithmic Equations

Solving logarithmic equations involves isolating the variable of interest and often requires the use of properties of logarithms to simplify the expressions. As seen in the step-by-step solution, after utilizing the properties of logarithms, we can often set up the equation in such a way that the solution becomes more apparent.

For example, in step 4, the task was to find the value of \( \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z} \). Once we have simplified our expressions using the logarithm properties, we can substitute the values back into this equation. Notably, the symmetry in the problem makes it even simpler, as each term in the sum turns out to have the same structure, effectively leading to a neat solution.

Solving logarithmic equations often follows a similar pattern: transforming the expression, using logarithmic identities to simplify, and then solving for the variable. By practicing this method, students can become adept at working through a variety of logarithmic problems. Remember to check for extraneous solutions, as sometimes the logarithmic operations impose restrictions on the variable that must be considered to arrive at a correct final answer.