SOLUTION: I am having problems figuring these out could you help please plot the graph f(x)=(1/2)^x f(x)=log2^x

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I am having problems figuring these out could you help please plot the graph f(x)=(1/2)^x f(x)=log2^x      Log On


   



Question 141087: I am having problems figuring these out could you help please
plot the graph
f(x)=(1/2)^x
f(x)=log2^x

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not sure exactly how you want to go about doing this. Are you allowed to use graphing calculators? That would be one way. Another way is by just plotting points. For the first one, just make a table by letting x = 0, x=1, x= 2, x=3, x=-1, x=-2, and x=-3. Find the corresponding values of f(x).
If x=0, f(0) = (1/2)^0 =1
If x=1, f(1) = (1/2)^1 =1/2
If x= 2, f(2) = (1/2)^2=1/4
If x=3, f(3) = (1/2)^3 =1/8
If x=-1, f(-1) = (1/2)^-1 =2
If x= -2, f(-2) = (1/2)^-2=4
If x=-3, f(-3) = (1/2)^-3 =8
graph+%28300%2C300%2C+-5%2C5%2C-10%2C10%2C%281%2F2%29%5Ex%29

For the second equation, you meant to write this: f%28x%29+=+log+%282%2Cx%29

By the definition of logarithms, y=+log%282%2Cx%29+ means 2%5Ey=x. So, make another table of values but in this case, start out with y=0, y=1, y=2, y=3, y=-1, y=-2, y=-3.
If y=0, then x = 2^0 =1
If y=1, then x = 2^1 =2
If y= 2, then x = 2^2=4
If y=3, then x = 2^3 =8
If y=-1, then x = 2^-1 =1/2
If y= -2, then x = 2^-2=1/4
If y=-3, then x = 2^-3 =1/8
The graph should look like this:
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C+log%282%2Cx%29%29

For additional explanation on the topic of Logarithms, see my own website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Take the second link on my homepage which is "Basic, Intermediate, and College Algebra: One Step at a Time", click on "College Algebra" and look for Chapter 4, Section 4.01 Logarithms. This is an explanation written for folks who didn't get math the first time!!
R^2