 My Math Forum Logarithmic Functions (Why are the graphs different?)
 User Name Remember Me? Password

 Pre-Calculus Pre-Calculus Math Forum

 August 13th, 2015, 02:34 PM #1 Member   Joined: Sep 2014 From: Toronto Posts: 43 Thanks: 0 Logarithmic Functions (Why are the graphs different?) Hey guys! I'm not quite sure how to answer this question. Any help is gladly appreciated. "The graphs of y= logx + log(2x) and y=log(2x^2) are different. Although the graphs are different, simplifying the first expression using the laws of logarithms produces the second expression. Explain why the graphs are different." Y= logx + log(2x) is on the right, Y = log(2x^2) is on the left. P.S. How do I delete a thread? August 13th, 2015, 02:47 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,693 Thanks: 2677 Math Focus: Mainly analysis and algebra The reason is that $\log x$ is defined only for $x \gt 0$. August 13th, 2015, 03:10 PM   #3
Member

Joined: Sep 2014
From: Toronto

Posts: 43
Thanks: 0

Quote:
 Originally Posted by v8archie The reason is that $\log x$ is defined only for $x \gt 0$.
Can you elaborate a bit more? Why is it only defined for such? August 14th, 2015, 03:56 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Are you saying that you are to answer this question without knowing what a "logarithm" is? Thanks from Zery August 14th, 2015, 09:56 AM   #5
Member

Joined: Sep 2014
From: Toronto

Posts: 43
Thanks: 0

Quote:
 Originally Posted by Country Boy Are you saying that you are to answer this question without knowing what a "logarithm" is?
Nvm I understood, just took me some time ha ha. Once again thank you. August 14th, 2015, 10:11 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Since the question has been raised (and I just have to put my oar in!), there are two common ways to define "ln(x)": The most common is to define the function $\displaystyle f(x)= e^x$ and then define ln(x) as the inverse function. Since any positive number to the x power maps the real numbers to the positive real numbers, its inverse, ln(x), maps the positive real numbers to the set of all real numbers and has domain x > 0. Another, in my opinion better, definition is $\displaystyle \ln(x)= \int_1^x (1/t)dt$. Since 1/x is not continuous for x < 0, that defines ln(x) only for x > 0. Last edited by skipjack; August 14th, 2015 at 05:05 PM. Tags functions, graphs, logarithmic Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Mr Davis 97 Algebra 1 May 3rd, 2014 10:02 AM strangepath Algebra 2 April 13th, 2013 01:50 PM pjlloyd100 Calculus 1 October 3rd, 2012 07:43 AM dphan26 Calculus 17 May 3rd, 2012 05:37 PM robasc Algebra 1 July 21st, 2008 07:23 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      