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August 13th, 2015, 03:34 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, 03:47 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 Math Focus: Mainly analysis and algebra 
The reason is that $\log x$ is defined only for $x \gt 0$.

August 13th, 2015, 04:10 PM  #3 
Member Joined: Sep 2014 From: Toronto Posts: 43 Thanks: 0  
August 14th, 2015, 04:56 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,879 Thanks: 766 
Are you saying that you are to answer this question without knowing what a "logarithm" is?

August 14th, 2015, 10:56 AM  #5 
Member Joined: Sep 2014 From: Toronto Posts: 43 Thanks: 0  
August 14th, 2015, 11:11 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,879 Thanks: 766 
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 06:05 PM. 

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functions, graphs, logarithmic 
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