GATE 2018 - Computer Science - Logarithm - 2 Marks
Duration: 4 min
This video lesson is available to enrolled students.
Inside: a video lesson and guided study material.
Module outline
- Quantitative Aptitude: Number System, Percentage, Profit and Loss, Ratio and Proportion (Ratios), Average, Time and Work, Speed Time and Distance, Simple Interest and Compound Interest, Mixture and Alligation, Data Interpretation, Permutation and Combination, Probability, Logarithm, Powers and Exponents (Surds and Indices), Age Problems, Divisibility Rules, Partnership, HCF LCM, Algebra, Pipe and Cistern
- Logical Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Analogy, Coding Decoding, Blood Relations, Directions (Direction Test), Syllogisms, Logical Sequence Of Words, Data Sufficiency, Seating Arrangements, Puzzle Solving (Puzzles), Calendar, Clock, Cubes & Dices, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Data Arrangement
- English Language: Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Vocabulary, Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Active Passive Voice (Voice), Direct and Indirect Speech, Cloze Test (Cloze Passage), Tenses, Subject Verb Agreement (Verb Noun Agreement), Preposition, Articles, Conjunction, Adverb
- Written English Test: English Writing Ability
- Coding: Coding Questions by Yash Sir, C Programming by Yash Sir, Programming in C++, Java, Python for Placements by Yash Sir
- Communication Round: Communication Assessment
- DBMS (Technical Interview): Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, File Organization & Indexing, Relational Algebra, Relational Calculus, Transaction Management, Concurrency Control, Database Recovery, ORDBMS, Database Security & Authorization, Query Processing & Optimization, Enhanced Data Models, Big Data Systems, NoSQL, Data Warehousing & Mining
- SQL (Technical Interview): SQL
- Operating Systems (Technical Interview): Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Deadlock, Threads & Process Creation, Memory Management, Virtual Memory, Disc Scheduling, File Management, Windows OS, Linux OS, Security, Distributed Systems, Virtual Machines
- Computer Networks (Technical Interview): Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware basics, Network Security, Mobile Technology, Cloud Computing and IoT, Cyber and Info Security
- OOPs (Technical Interview): Object Oriented Programming (OOPS)
- HR Interview: MR & HR Interview Preparation
AI summary & chapters
AI Summary
An AI-generated summary of this video lecture.
This video is a tutorial on solving a mathematical problem from the GATE CS 2018 exam. The problem presents three equations: p^(-x) = 1/q, q^(-y) = 1/r, and r^(-z) = 1/p, and asks for the value of the product xyz. The instructor, Yash Jain Sir, uses the logarithmic identity log(a^m) = m * log(a) to solve the problem. He takes the logarithm of both sides of each equation, which allows him to express x, y, and z in terms of logarithms of p, q, and r. He then multiplies these three expressions together, and through algebraic simplification, he shows that the product xyz equals 1. The video concludes with the instructor confirming the final answer.
Chapters
0:00 – 2:00 00:00-02:00
The video opens with a title card for 'GATE APTITUDE LOGARITHMS'. The main content begins with a problem displayed on a whiteboard: 'Q. If pqr ≠ 0 and p^(-x) = 1/q, q^(-y) = 1/r, r^(-z) = 1/p, what is the value of the product xyz? (GATE CS 2018) (2 MARKS)'. The instructor, Yash Jain Sir, introduces the problem and the four multiple-choice options: A. -1, B. 1/pqr, C. 1, D. pqr. He then begins to solve the problem by taking the logarithm of both sides of the first equation, p^(-x) = 1/q, which he writes as log(p^(-x)) = log(1/q). He applies the logarithmic rule log(a^m) = m * log(a) to get -x * log(p) = log(1/q). He then uses the rule log(1/a) = -log(a) to simplify the right side to -log(q). This gives him the equation -x * log(p) = -log(q), which he simplifies to x * log(p) = log(q). He then solves for x, writing x = log(q) / log(p). He proceeds to do the same for the second equation, q^(-y) = 1/r, to find y = log(r) / log(q). He then begins to work on the third equation, r^(-z) = 1/p, writing log(r^(-z)) = log(1/p) and simplifying to -z * log(r) = -log(p), which gives z = log(p) / log(r).
2:00 – 3:34 02:00-03:34
The instructor now has the three expressions for x, y, and z: x = log(q) / log(p), y = log(r) / log(q), and z = log(p) / log(r). He states that the goal is to find the product xyz. He writes the product as (log(q) / log(p)) * (log(r) / log(q)) * (log(p) / log(r)). He then performs the multiplication, showing that the terms cancel out: the log(q) in the numerator of the first fraction cancels with the log(q) in the denominator of the second, the log(r) in the numerator of the second cancels with the log(r) in the denominator of the third, and the log(p) in the numerator of the third cancels with the log(p) in the denominator of the first. This leaves him with 1. He concludes that the product xyz is equal to 1. He then points to the multiple-choice options and confirms that the correct answer is C. 1. The video ends with a 'THANKS FOR WATCHING' screen.
The video presents a clear, step-by-step solution to a GATE exam problem using logarithmic properties. The core of the solution is the application of the logarithmic identity log(a^m) = m * log(a) to transform the given exponential equations into a form where the variables x, y, and z can be isolated. The final step is a clever algebraic simplification of the product xyz, where the intermediate terms cancel out, leading to the final answer of 1. The instructor's methodical approach demonstrates a powerful technique for solving systems of equations involving exponents.