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Chapter 5: Problem 18

Prove that \(\frac{\log _{a} n}{\log _{a b} n}=1+\log _{a} b\)

Short Answer

Expert verified
The proof that \(\frac{\log _{a} n}{\log _{a b} n}=1+\log _{a} b\) is done by using the properties of logarithms, particularly the change of base formula and the logarithm product rule. The final result is \( \log_a n + \log_a n \cdot \log_a b = 1+\log_a b \).

Step by step solution

01

Change base formula

Start by applying the change of base formula to the denominator. The change of base formula states that \(\log_b a = \frac{\log_d a}{\log_d b}\). Thus, \(\log_{ab} n = \frac{\log_a n}{\log_a ab}\). So, the original expression can be rewritten as \(\frac{\log_a n}{\frac{\log_a n}{\log_a ab}} = \frac{\log_a n}{1/\log_a ab}\)
02

Simplify fraction

Next, simplify the fraction by multiplying the numerator and denominator by \(\log_a ab\), that gives \( \frac{\log_a n \cdot \log_a ab}{1} = \log_a n \cdot \log_a ab\).
03

Rewriting expression using product rule

Then, use the log product rule to rewrite the expression further. The product rule of logs states that \(\log_a mn = \log_a m + \log_a n\). Here, \(\log_a n \cdot \log_a ab = \log_a n(1+\log_a b)\).
04

Simplify expression

Finally, simplify the expression by distributing \(\log_a n\) to get \( \log_a n + \log_a n \cdot \log_a b = 1+\log_a b \). Thus, proving the equation.

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Key Concepts

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

Logarithm Properties
To fully understand the solution to the provided logarithmic exercise, one must first grasp the fundamental properties of logarithms. Logarithms, the inverse operations of exponentiation, have properties that simplify complex expressions into manageable terms.

  • Product Rule: The logarithm of a product is equal to the sum of logarithms of the individual factors, represented as \(\log_b(mn) = \log_b m + \log_b n\).
  • Quotient Rule: The logarithm of a quotient is the difference of the logarithms of the numerator and the denominator, shown as \(\log_b(\frac{m}{n}) = \log_b m - \log_b n\).
  • Change of Base Formula: Logarithms with different bases can be converted using the formula \(\log_b a = \frac{\log_d a}{\log_d b}\), allowing for base conversion and simplification of the expression.
  • Power Rule: The logarithm of a number raised to an exponent is equal to the exponent multiplied by the logarithm of the number, which is written as \(\log_b(m^k) = k\cdot\log_b m\).
These properties are crucial in logarithmic manipulation and solving equations involving logs, forming a foundational part of higher mathematics education, including preparation for competitive exams like the IIT JEE.
Trigonometry Problem Solving
Solving trigonometry problems involves understanding the relationships between the angles and sides of triangles and the properties of trigonometric functions. Steps to approach a trigonometry problem effectively include:

  • Identifying the right trigonometric ratios and functions to use based on the given information.
  • Applying theorems and properties of triangles, such as the Pythagorean theorem and triangle sum theorem.
  • Utilizing identities like the sine and cosine rules or the tangent rule for non-right triangles.
  • Manipulating expressions using angle sum and difference identities.
  • Converting between different trigonometric forms if necessary, such as from sine to cosine.
Adequate practice with trigonometry problem solving helps in building intuition and skillset required for tackling complex problems often encountered in competitive exams and in various fields of science and engineering.
IIT JEE Trigonometry
The subject of trigonometry in the Indian Institutes of Technology Joint Entrance Examination (IIT JEE) is known for its rigor and the depth of understanding it requires. Aspiring engineers must have a strong command over trigonometry to excel in this exam. Key areas one must master include:

  • Trigonometric functions, their graphs, and transformation.
  • In-depth knowledge of trigonometric identities and equations.
  • Properties of triangles, including the solution of triangles.
  • Application of trigonometric functions to solve problems involving heights and distances.
  • Understanding inverse trigonometric functions and their properties.
Thorough preparation focusing on these concepts, alongside consistent practice solving varied and complex problems, will enable students to approach the trigonometry section of the IIT JEE with confidence and a strong likelihood of success.

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Most popular questions from this chapter

If \(a, b, c\) are in G.P. then \(\log _{2016} a, \log _{2016} b, \log _{2016} c\) are in (a) G.P. (b) A.P. (c) H.P. (d) A.G.P.

The value of \(\left(\frac{1}{\log _{3} \pi}+\frac{1}{\log _{4} \pi}\right)\) lies between (a) \((1,2)\) (b) \((2,3)\) (c) \((3,4)\) (d) \((0,1)\)

\(\log _{(1 / 2)}\left(\log _{5}\left(\log _{2}\left(x^{2}-6 x+40\right)\right)\right)>0\)

Find the value of (i) \(\log _{10} \tan 40^{\circ}+\log _{10} \tan 41^{\circ}+\log _{10} \tan 42^{\circ}\) \(+\ldots \ldots+\log _{10} \tan 50^{\circ}\) (ii) \(\log _{10} \tan 1^{\circ}+\log _{10} \tan 2^{\circ}+\log _{10} \tan 3^{\circ}\) \(+\ldots \ldots \ldots . .+\log _{10} \tan 89^{\circ}\) (iii) \(\log _{3} 4 \cdot \log _{4} 5 \cdot \log _{5} 6 \cdot \log _{6} 7 \cdot \log _{7} 8 \cdot \log _{8} 9\)

\(\log _{(1 / 2)} x>\log _{(1 / 3)} x\)
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