Intermediate algebra solving logarithm

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becky0307

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Feb 10, 2010
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Hi,
I'm trying to do this problem and I need to graph it as well. I think I did it all wrong after reading my book again.
The problem is;
find 3 values of x that are less than -1, 3 that are between 0 and -1 and x at -1. use the resulting pairs to plot the graph. state the equation of the line asymptotic to the graph.

Equation is: y=-log3.5 (-x)
I was calculating -log 3.5 and then multiplying by (-(-x)).
should I be restating the equation in the form of x=b^y ?
then if I do that would I have -3.5^y=-x
 
becky0307 said:
Hi,
I'm trying to do this problem and I need to graph it as well. I think I did it all wrong after reading my book again.
The problem is;
find 3 values of x that are less than -1, 3 that are between 0 and -1 and x at -1. use the resulting pairs to plot the graph. state the equation of the line asymptotic to the graph.

Equation is: y=-log3.5 (-x)
I was calculating -log 3.5 and then multiplying by (-(-x)).
should I be restating the equation in the form of x=b^y ?
then if I do that would I have -3.5^y=-x
Click to expand...

Is your function:

\(\displaystyle y = - LOG(3.5) * (-x)\)

or

\(\displaystyle y = - LOG_{3.5}(-x)\)

Those are completely different functions
 
First, you should probably convert the equation to a base 10 logarithm since most calculators can't calculate logarithms except base ten and base e. You can find a simple explanation of how to do that here: http://mathforum.org/library/drmath/view/55565.html. The result is:

\(\displaystyle y = -\frac{LOG_{10}-x}{LOG_{10}3.5}\)

Then, let's choose some points to look up (I'll fill in the answers for a few so you can make sure you are calculating it right).

x < -1
(-4, y)
(-3, -0.88)
(-2, y)

x = -1
(-1, y)

-1 < x < 0
(-0.5, 0.55)
(-0.1, y)
(-0.01, y)

If you're not sure what the answer is, I also suggest you try x = 0.

Then plot the points on a graph and draw a line through them. If you need to, choose more points to help you understand the shape of the graph.
 
Thanks, I found that same example in my book, was looking right at it and never put it together!
I think I got it now!
 
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