Logarithm Equation Question 2 [Checking validity of answers in logarithms equations]

Solve the equation

\lg(3a+2)+ lg(5a-3)=1+ 2lg (a)

Ans : a = 1 or -6/5

Question : Do we need to include -6/5

Contributed by YC

It is important for you to check if your answers are valid i.e are there any answers you would need to reject?
This is what I called checking for validity of answers.

Related Post

'O' levels & Sec 3 Math & Chemistry Holiday Programmes 2010 (Revision & Headstart) UPDATE: Early Bird registration is extended to 6 Dec 09, 2359 hours (Singapore time) Credit: Art Institute of Portland Most Secondary two st...
Logarithm Equation Question 1 I met a student and she passed me this paper with this Lg question on it, asking me on the solution for solving this equation : Coincidentally, I w...
Linear Law,Just 1 Strategy Makes A Grade Difference What is Linear Law? It is a tool which will allow you to transform non-straight line equation to straight line equation so that you can plot a stra...
O level A-Math: Fatal Mistake In Logarithm Equation You Must Aviod This is a common mistake I notice in my O level A-Math essential concepts revision workshop. Solve the following equation: First, you must i...
Top 7 Commonly Made Mistakes in Logarithm How many of you would like to learn Math faster? And how many of you would like to score more marks for Math? Do you know that one of the fastest ...

Hi, I'm Ai Ling Ong. I enjoy coaching students who have challenges with understanding and scoring in 'O' Level A-Maths and E-Maths. I develop Math strategies, sometimes ridiculous ideas to help students in understanding abstract concepts the fast and memorable way. I write this blog to share with you the stuff I teach in my class, the common mistakes my students made, the 'way' to think, analyze... If you have found this blog post useful, please share it with your friends. I will really appreciate it! :)

5 Responses to Logarithm Equation Question 2 [Checking validity of answers in logarithms equations]

  1. Clearly the domain (admissible values of x for the argument of the function f(x)=lg(x)) of f(x) is (0, \infty). If a=-6/5, then 3a+2 is negative; therefore x=-6/5 is not admissible into the domain of function f(x).


Leave a reply