SOLUTION: I have another question I am stuck on. 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.

Question 315194: I have another question I am stuck on. 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. State the equation of the line asymptotic to the graph.
y=log2(x+2)
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I have another question I am stuck on. 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. State the equation of the line asymptotic to the graph.
y=log2(x+2)
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I think you mean log base 2 ?

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Calculators don't have logs base 2, so you use the base change formula

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You sub the values for x, and find y.
It says 3 values > -1 and 3 values between -1 and 0. The values between -1 and 0 will be greater than -1 ??? That's redundant.
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x y
-1 0
-0.8 0.263034406
-0.6 0.485426827
-0.4 0.678071905
-0.2 0.847996907
0 1
0.2 1.137503524
0.4 1.263034406
0.6 1.378511623
0.8 1.485426827
1 1.584962501
I used Excel
email me at moral loophole@aol.com if you want a graph
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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. State the equation of the line asymptotic to the graph.
y=log2(x+2)
If x = -1, y = log2(1) = 0
If x = 0, y = log2(2) = 1
If x = 2, y = log2(4) = 2
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Vertical asymptote at x = -2
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If x = -1/4 , y = log2(7/4) = log2(7) - 2
If x = -1/2 , y = log2(3/2) = log2(3) - 1
If x = -3/4 , y = log2(5/4) = log2(5) - 2
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Cheers,
Stan H.

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