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Question 212626: I want to make sure I did this right. And I was wondering what the graph looks like. I think I did something wrong.
Evaluate the logarithmic equation for three values of x that are greater than 1, three values of x that are between 0 and 1, and at x=1. Show your work. Use the resulting ordered pairs to plot the graph; submit the graph via the Dropbox. State the equation of the line asymptotic to the graph (if any).
Question: y = log4 (x + 4)
y = log4 (x + 4)
y=log4(1+4)
y=5ln/4ln
x=1 y=1.2
y=log4(2+4)
y=6ln/4ln
x=2 y= 1.3
y=log4(3+4)
y=7ln/4ln
x=3 y=1.4
y=log4(4+4)
y=8ln/4ln
x=4 y=1.5
y=log4(.25+4)
y=4.25ln/4ln
x=.25 y=1.04
y=log4(.5+4)
y=4.5ln/4ln
x=.5 y=1.1
y=log4(.75+4)
y=4.75ln/4ln
x=.75 y=1.12
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I want to make sure I did this right. And I was wondering what the graph looks like. I think I did something wrong.
Evaluate the logarithmic equation for three values of x that are greater than 1, three values of x that are between 0 and 1, and at x=1. Show your work. Use the resulting ordered pairs to plot the graph; submit the graph via the Dropbox. State the equation of the line asymptotic to the graph (if any).
Question: y = log4 (x + 4)
y = log4 (x + 4)
y=log4(1+4)
y=5ln/4ln
x=1 y=1.2
y=log4(2+4)
y=6ln/4ln
x=2 y= 1.3
y=log4(3+4)
y=7ln/4ln
x=3 y=1.4
y=log4(4+4)
y=8ln/4ln
x=4 y=1.5
y=log4(.25+4)
y=4.25ln/4ln
x=.25 y=1.04
y=log4(.5+4)
y=4.5ln/4ln
x=.5 y=1.1
y=log4(.75+4)
y=4.75ln/4ln
x=.75 y=1.12
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I'm afraid there is some misunderstanding:
For example:
y = log(base 4) (x+4)
If x = 1,
y = log(base 4) (5)
y = ln(5)/ln(4) = 1.1609
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Cheers,
Stan H.
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