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Chapter 12: Problem 43

The value of $$ \left(\left(\log _{2} 9\right)^{2}\right)^{\frac{1}{\log _{2}\left(\log _{2} 9\right)}} \times(\sqrt{7})^{\frac{1}{\log 4} 7} $$ is

Short Answer

Expert verified
\(7 \log_{2}9\) or approximately 22.458.

Step by step solution

01

Simplify the Exponent in First Term

Recognize that raised to the power of its own reciprocal log base is equivalent to the base itself. Thus, simplify the expression \(\left(\log _{2} 9\right)^{2}\right)^{\frac{1}{\log _{2}\left(\log _{2} 9\right)}}\) to \(\log _{2} 9\).
02

Simplify the Exponent in Second Term

Understand that any number raised to the power of \(\frac{1}{\log_{} }\) equals the base. Apply it to \(\sqrt{7}\) raised to \(\frac{1}{\log 4} 7\), and recognize that \(\sqrt{7} = 7^{\frac{1}{2}}\), resulting in: \(7^{\frac{1}{2} \cdot \frac{2}{\log 7}} = 7^{\frac{1}{\log 7}} = 7\).
03

Multiply the Simplified Terms

Multiply the simplified terms \(\log _{2} 9\) and 7 together to get the value of the entire expression: \(\log _{2} 9 \times 7\).
04

Calculate the Numerical Value

Use logarithm properties to evaluate \(\log _{2} 9\). This value does not simplify to a whole number, so we would only need to represent it as \(\log _{2} 9\) or use a calculator to approximate. Multiplying \(\log _{2} 9\) by 7 gives us the final answer in either exact (logarithmic) or approximate form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithm Properties
Understanding the properties of logarithms is essential when working with logarithmic expressions. Logarithms, the inverse operations of exponentiation, have characteristics that can simplify complex expressions. Let's examine key properties that are commonly used:
  • Product Property: The log of a product is the sum of the logs: \[ \log_b(mn) = \log_b(m) + \log_b(n) \]
  • Quotient Property: The log of a quotient is the difference of the logs: \[ \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \]
  • Power Property: The log of a power is the exponent times the log: \[ \log_b(m^n) = n\cdot\log_b(m) \]
  • Change of Base Formula: Allows conversion of logs to a different base: \[ \log_b(m) = \frac{\log_k(m)}{\log_k(b)} \] where \( k \) is any positive number.
  • Reciprocal Property: Involves raising a log term to the power of its own reciprocal base, resulting in the base itself: \[ (\log_b(m))^n \cdot \frac{1}{\log_b(m)^n} = m \]
These properties are instrumental in transforming complex logarithmic expressions into simpler forms. For example, in our exercise, the Reciprocal Property is used to simplify the first term to \(\log_2 9\), since the exponent was the reciprocal of the base logarithm.
Simplifying Exponents
When simplifying exponents, it's crucial to apply the fundamental rules of exponents to condense expressions. Some often-used exponent rules include:
  • Product of Powers: Multiplying same base terms you add the exponents: \[ a^m \cdot a^n = a^{m+n} \]
  • Power of a Power: Raising an exponent to another exponent, you multiply the exponents: \[ (a^m)^n = a^{m\cdot n} \]
  • Power of a Product: Distribute the exponent over a product inside brackets: \[ (ab)^m = a^m \cdot b^m \]
  • Negative Exponent: Represents reciprocal, where \( a^{-n} = \frac{1}{a^n} \)
  • Fractional Exponents: A fractional exponent indicates roots, where \( a^{\frac{1}{n}} = \sqrt[n]{a} \) and \( a^{\frac{m}{n}} \) equates to \( \sqrt[n]{a^m} \).
By using these rules, complex expressions are made manageable. For instance, in the given exercise, we applied the rule of Fractional Exponents to recognize that \( \sqrt{7} = 7^{\frac{1}{2}} \). This was then used in conjunction with the reciprocal base property to narrow down the expression to just \( 7 \), greatly simplifying the original problem.
JEE Advanced Mathematics
The Joint Entrance Examination (JEE) Advanced is a highly competitive test for aspirants seeking admission to the Indian Institutes of Technology (IITs). Mathematics is a significant part of the exam, testing students' understanding of advanced concepts. Key topics within JEE Advanced Mathematics include Algebra, Calculus, Trigonometry, Geometry, and Probability. Mastery over these topics requires a deep understanding of fundamental concepts such as logarithms and exponents, which come up frequently in problems requiring analytical problem-solving skills.

Students preparing for JEE Advanced should focus on clarifying their concepts and application skills. Problems often involve multiple steps and the manipulation of expressions using properties of logarithms and exponents, as seen in our exercise. Strong problem-solving skills, coupled with an understanding of core mathematical principles, can give students an edge in this rigorous examination. As such, exercises similar to the one we discussed not only test a student's ability in simplification but also prepare them for the types of complex questions they may encounter in their JEE Advanced Mathematics exam.

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Most popular questions from this chapter

The solubility of a salt of weak acid \((\mathbf{A B})\) at \(\mathrm{pH} 3\) is \(\mathbf{Y} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\). The value of \(\mathbf{Y}\) is (Given that the value of solubility product of \(\mathbf{A B}\left(K_{s p}\right)=2 \times 10^{-10}\) and the value of ionization constant of \(\mathbf{H B}\left(K_{a}\right)=1 \times 10^{-8}\) )

Two infinitely long straight wires lie in the \(x y\) -plane along the lines \(x=\pm R\). The wire located at \(x=+R\) carries a constant current \(I_{1}\) and the wire located at \(x=-R\) carries a constant current \(I_{2} .\) A circular loop of radius \(R\) is suspended with its centre at \((0,0, \sqrt{3} R)\) and in a plane parallel to the \(x y\) -plane. This loop carries a constant current \(I\) in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the \(+\hat{\jmath}\) direction. Which of the following statements regarding the magnetic field \(\vec{B}\) is (are) true? (A) If \(I_{1}=I_{2}\), then \(\vec{B}\) cannot be equal to zero at the origin \((0,0,0)\) (B) If \(I_{1}>0\) and \(I_{2}<0\), then \(\vec{B}\) can be equal to zero at the origin \((0,0,0)\) (C) If \(I_{1}<0\) and \(I_{2}>0\), then \(\vec{B}\) can be equal to zero at the origin \((0,0,0)\) (D) If \(I_{1}=I_{2}\), then the \(z\) -component of the magnetic field at the centre of the loop is \(\left(-\frac{\mu_{0} I}{2 R}\right)\)

Consider an ionic solid MX with \(\mathrm{NaCl}\) structure. Construct a new structure \((\mathbf{Z})\) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance. (i) Remove all the anions \((\mathbf{X})\) except the central one (ii) Replace all the face centered cations ( \(\mathbf{M}\) ) by anions \((\mathbf{X})\) (iii) Remove all the corner cations \((\mathbf{M})\) (iv) Replace the central anion \((\mathbf{X})\) with cation \((\mathbf{M})\) The value of \(\left(\frac{\text { number of anions }}{\text { number of cations }}\right)\) in \(\mathbf{Z}\) is

For the electrochemical cell, $$ \mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s}) $$ the standard emf of the cell is \(2.70 \mathrm{~V}\) at \(300 \mathrm{~K}\). When the concentration of \(\mathrm{Mg}^{2+}\) is changed to \(x \mathrm{M}\), the cell potential changes to \(2.67 \mathrm{~V}\) at \(300 \mathrm{~K}\). The value of \(x\) is (given, \(\frac{F}{R}=11500 \mathrm{~K} \mathrm{~V}^{-1}\), where \(F\) is the Faraday constant and \(R\) is the gas constant, \(\ln (10)=2.30)\)

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle \(60^{\circ}\) with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is \((2-\sqrt{3}) / \sqrt{10} s\), then the height of the top of the inclined plane, in metres, is Take \(g=10 \mathrm{~ms}^{-2}\).
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