SOLUTION: Can you explain me how do I find the asymptotes of the functions f(x)=3^x and g(x) = log5(2x).

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Question 132970: Can you explain me how do I find the asymptotes of the functions f(x)=3^x and
g(x) = log5(2x).

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+3%5Ex
First, x+=+0 is not an asymptote because f%280%29+=+3%5E0
and 3%5E0+=+1. f%28x%29 changes smoothly on either side of x+=+0
3%5E0.01+=+1.011 and 3%5E%28-.01%29+=+.989, so the plot is
f%28x%29 . . x
.989 . . -0.01
1 . . 0
1.011 . . 0.01
The smooth changes in both f%28x%29 and x means there is
no asyptote
-------------
As x becomes infinitely large in the + direction, f%28x%29
becomes larger at an even greater rate without any bounds.
As x becomes larger in the negative direction, however,
f%28x%29 gets closer and closer to f%28x%29+=+0, but never
becomes negative. There is no way 3%5Ex can possibly be negative.
f%28x%29+=+0 is the definition of the x-axis, so the negative
x-axis is an asyptote
--------------------------
g(x) = log5(2x)
If I say y+=+g%28x%29, I can rewrite the equation as
5%5Ey+=+2x
Is x+=+0 an asymptote? If x+=+0, g%28x%29 approaches
negative infinity. There is no way that 2x can ever be
negative since 5%5Ey can never be negative. x+=+0 is the
definition of the y-axis, so the y-axis is an asymptote
When x becomes very large in the + direction, g%28x%29
also becomes very large, just not at as fast a rate.
I'll plot the curves
+graph%28+600%2C+600%2C+-5%2C+3%2C+-3%2C+3%2C+log%285%2C2x%29%2C3%5Ex%29+