Transform to a log

[justan4cat]

I'm in my last week of this online class. You guys have been sp helpful! I'm sorry if some of my questions were dumb questions, but I had nobody else to ask. Thanks for not poking fun. I have 2 problems left and I'm done with this course. This one, I'm a bit stuck on the second part. Here is what I have now.

Evaluate the exponential equation for three positive values of x, three negative values of x, and at x=0. Transform the second expression into the equivalent logarithmic equation; and 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 of each function.
Y = (5/2)^x, X = (5/2)^y

Y= 2^-5x X=1,1.25,1.5, -1,-1.25,-1.5,0
(1,.0313),(1.25,.0131),(1.5,.0055)
(-1,2),(-1.25,.1768),(-1.5,.3536),(0,1) same thing for X in the second equation, just reversed.

Then comes the part where I have to "transform" the x=(5/2)^y to a logarithmic equation and pick the pairs again. I looked at all my notes, and can't find one that fits this. What would the equation be?

 

[Deleted member 4993]

I'm in my last week of this online class. You guys have been sp helpful! I'm sorry if some of my questions were dumb questions, but I had nobody else to ask. Thanks for not poking fun. I have 2 problems left and I'm done with this course. This one, I'm a bit stuck on the second part. Here is what I have now.

Evaluate the exponential equation for three positive values of x, three negative values of x, and at x=0. Transform the second expression into the equivalent logarithmic equation; and 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 of each function.
Y = (5/2)^x, X = (5/2)^y

Y= 2^-5x <<< Where did this come from? It is not equivalent to Y = (5/2)^x,
X=1,1.25,1.5, -1,-1.25,-1.5,0
(1,.0313),(1.25,.0131),(1.5,.0055)
(-1,2),(-1.25,.1768),(-1.5,.3536),(0,1) same thing for X in the second equation, just reversed.

Then comes the part where I have to "transform" the x=(5/2)^y to a logarithmic equation and pick the pairs again. I looked at all my notes, and can't find one that fits this. What would the equation be?

 

[justan4cat]

The example in my book begins with:

y=f(x)=(1/2)[sup:dgohqule]x[/sup:dgohqule]=(2[sup:dgohqule]-1[/sup:dgohqule])[sup:dgohqule]x[/sup:dgohqule]=2[sup:dgohqule]-x[/sup:dgohqule]

I thought that was the first step. This is why I need you guys, when I just go by my book, I get stuff wrong. Today is my last day of this class, so if I get this problem or not, I have to turn something in before midnight. I really would appreciate your help so I at least get a few points towards this one.

 

[Denis]

Transform the second expression into the equivalent logarithmic equation; and 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.
X = (5/2)^y

OK; RULE: if a^p = b then p = log(b) / log(a) : tattoo that on your wrist!
So if (5/2)^y = x, then:
y = log(x) / log(5/2) : got that??

Say you make x = 5; then:
y = log(5) / log(5/2) = 1.75647......

Hope that helps.....

NOTE: also, remember that log(1) = 0