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August 13th, 2015, 02:34 PM   #1
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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?
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August 13th, 2015, 02:47 PM   #2
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The reason is that $\log x$ is defined only for $x \gt 0$.
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August 13th, 2015, 03:10 PM   #3
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Quote:
Originally Posted by v8archie View Post
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?
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August 14th, 2015, 03:56 AM   #4
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Are you saying that you are to answer this question without knowing what a "logarithm" is?
Thanks from Zery
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August 14th, 2015, 09:56 AM   #5
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Quote:
Originally Posted by Country Boy View Post
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.
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August 14th, 2015, 10:11 AM   #6
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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.
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