Question

If \(3^{x}=4^{x-1}\), then \(x=\) (A) \(\frac{2 \log _{3} 2}{2 \log _{3} 2-1}\) (B) \(\frac{2}{2-\log _{2} 3}\) (C) \(\frac{1}{1-\log _{4} 3}\) (D) \(\frac{2 \log _{2} 3}{2 \log _{2} 3-1}\)

Short answer

The value of x is given by option (A) \(\frac{2 \log_{3} 2}{2 \log_{3} 2 - 1}\).

Step-by-step solution

Take the logarithm of both sides

To solve the equation, first take the natural logarithm (or logarithm with any base) of both sides to utilize the properties of logarithms. This is done to express the exponents in a form that can be solved algebraically.

Apply the power rule of logarithms

Use the power rule of logarithms, which states that \(\log(a^b) = b \log(a)\), to move the exponents in front of the logarithms. This yields: \(x \log(3) = (x - 1) \log(4)\).

Expand and rearrange the terms

Expand the right-hand side of the equation and start rearranging the terms to isolate the variable \(x\). This results in \(x \log(3) = x \log(4) - \log(4)\), and then \(x \log(3) - x \log(4) = -\log(4)\).

Factor out the common variable

Factor out \(x\) from the left-hand side to simplify the equation: \(x(\log(3) - \log(4)) = -\log(4)\).

Solve for the variable \(x\)

Divide both sides of the equation by the factor \((\log(3) - \log(4))\) to solve for \(x\): \(x = -\frac{\log(4)}{\log(3) - \log(4)}\).

Simplify the expression

Using the change of base formula \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\), the expression for \(x\) simplifies to \(x = \frac{2 \log(2)}{2\log(2) - 1}\), which corresponds to the option (A) \(\frac{2 \log_{3} 2}{2 \log_{3} 2 - 1}\).

Key concepts

Logarithm Properties

Logarithms are incredibly useful in solving equations where the variable is an exponent, such as in the given exercise. Understanding the properties of logarithms is essential to manipulating and solving logarithmic equations. One fundamental property is the power rule, which states that \(\log(a^b) = b \log(a)\). This property allows you to take an exponent and move it in front of the logarithm, converting a multiplicative process into an additive one.

Another key property is the change of base formula. It allows us to switch from one base of a logarithm to another, which is especially useful when simplifying expressions or comparing logarithms with different bases. This is expressed as \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\), where \(b\) and \(c\) are bases, and \(a\) is the argument of the logarithm. These properties are crucial for problem-solving in advanced mathematics, as illustrated in the solution of the exercise.

Exponential Equations

Exponential equations are those where the unknown variable appears in the exponent, such as \(3^{x}=4^{x-1}\) from our exercise. To solve these equations, one often needs to apply logarithms to both sides; this renders the exponents accessible for algebraic manipulation. The process of taking logarithms helps to linearize the equation, making it possible to isolate the variable and solve for it.

In the context of JEE Advanced Mathematics, solving exponential equations is a vital skill. Competency in these equations lays the groundwork for further topics in calculus, differential equations, and beyond. Students should practice solving various exponential equations to gain familiarity with different strategies, such as equating bases or applying logarithms, as this proficiency can be a significant advantage in a competitive exam setting.

JEE Advanced Mathematics

The Joint Entrance Examination (JEE) Advanced is known for its challenging mathematical problems that test a student's conceptual understanding and application skills. Problems from logarithmic and exponential equations are a staple in this high-level exam. These topics not only require memorization of the formulas and properties but also a deep understanding of how to apply them in unconventional ways.

Students preparing for the JEE Advanced should focus on developing a strong grasp of fundamental concepts and practicing a variety of problems to enhance their problem-solving abilities. Emphasizing clarity of concept, as seen in the given step-by-step solution for a logarithmic equation, is critical. This approach aids in understanding the nature of the problem and leads to effective strategies to arrive at the correct answer efficiently. Practicing such problems helps in honing the skills needed to excel in the mathematics section of the JEE Advanced and similar competitive examinations.