Problem 28 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Number of solutions of \(\log _{2}\left(9-2^{x}\right)=10^{\log _{s}(3-x)}\) is (A) 1 (B) 2 (C) 3 (D) None of these

Step-by-Step Solution

Answer

(A) 1

Step 1: Understand the Problem

Our goal is to find the number of solutions for the equation \( \log_{2}(9-2^{x})=10^{\log_{s}(3-x)} \). We will break down the equation and find conditions under which it is valid.

Step 2: Simplify the Right Side

The right side, \( 10^{\log_{s}(3-x)} \), indicates a logarithmic equation with an unspecified base \( s \). For simplicity, let's assume \( s = 10 \), which converts \( 10^{\log_{10}(3-x)} \) into \( 3-x \). Thus, the equation simplifies to \( \log_{2}(9-2^{x}) = 3-x \).

Step 3: Convert Logarithmic Expression

To solve \( \log_{2}(9-2^{x}) = 3-x \), rewrite it as an exponential equation: \( 9 - 2^{x} = 2^{3-x} \). This expression tells us the condition that must hold true for the equality to be maintained.

Step 4: Solve the Equation

We have \( 9 - 2^{x} = 2^{3-x} \). Assume \( x = a \). Substituting, we obtain: \( 2^{a} + 2^{3-a} = 9 \). Now try possible values of \( a \).

Step 5: Solve for Specific Values

Check the value of \( x \). If \( x = 2 \), then \( 2^2 + 2^{3-2} = 4 + 2 = 6 \), not equal to 9. If \( x = 1 \), \( 2^1 + 2^{3-1} = 2 + 4 = 8 \), also not equal to 9. When \( x = 0 \), \( 2^0 + 2^{3-0} = 1 + 8 = 9 \). Thus, \( x = 0 \) is a solution.

Step 6: Check for Additional Solutions

Confirm if there are other possible solutions using any other method or numerical value checks. By testing, \( x = 0 \) is the only value that satisfies the equation under the assumption of base 10 for the right side logarithm.

Key Concepts

Exponential Equations

Exponential equations are interesting problems in mathematics where the variable appears as an exponent. This means that solving these equations can be a bit tricky since you're not just dealing with straightforward numbers but powers. A classic form of an exponential equation is seen when you rewrite a logarithmic equation into this format. For example, consider the expression we derived: \( 9 - 2^{x} = 2^{3-x} \). It shows how exponential terms can be balanced in an equation. To solve such equations, it is common to perform substitutions and transformations to simplify the exponential forms. The essence of solving exponential equations lies in:

• Refining the base to simplify comparisons
• Using properties of exponents effectively
• Testing values systematically to find solutions

By analyzing exponential equations carefully, you gain the skills to manipulate and solve them, which is a powerful tool in algebra.

Number of Solutions

Determining the number of solutions to an equation requires careful analysis of terms and conditions. In the context of our problem, this means checking each possible scenario where the equation holds true. When solving the equation \( 9 - 2^{x} = 2^{3-x} \), we tested different values for \( x \) to determine when the left-side and right-side equal. To make sure every potential solution is accounted for, consider:

• Checking obvious and simple values first, such as whole numbers
• Evaluating special cases where the equation simplifies significantly
• Recognizing when to stop searching based on logical constraints

In this exercise, we confirmed through numeric testing that \( x = 0 \) is the only solution. Having just a single value that made the equation true indicates there is exactly one solution.

Logarithm Properties

Logarithm properties are an essential toolset when dealing with equations that involve logarithms, such as the original logarithmic equation given. These properties help in transforming and simplifying expressions so that they can be solved more easily. Key logarithm properties include:

• The Power Rule: \( \log_{b}(a^c) = c\log_{b}(a) \)
• The Change of Base Formula: \( \log_{b}(a) = \frac{\log_{k}(a)}{\log_{k}(b)} \)
• The Inversion Rule: \( b^{\log_{b}(a)} = a \)

In our exercise, we applied a crucial logarithmic conversion when simplifying \( 10^{\log_{s}(3-x)} \) by assuming the base \( s \) to be 10. This leverages the inversion rule to simplify the expression directly to \( 3-x \). Using logarithm properties skillfully helps in tackling complex mathematical problems by breaking them down into simpler parts.