SOLUTION: Write an exponential function for a graph that includes point (1, 0.84) and (2,1.008). I've gotten as far as 0.84/b=a , and 1.008=(.84/b)*b^2, but I'm stuck there.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Write an exponential function for a graph that includes point (1, 0.84) and (2,1.008). I've gotten as far as 0.84/b=a , and 1.008=(.84/b)*b^2, but I'm stuck there.      Log On


   

Question 571114: Write an exponential function for a graph that includes point (1, 0.84) and (2,1.008). I've gotten as far as 0.84/b=a , and 1.008=(.84/b)*b^2, but I'm stuck there.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I would say the function is y=a%2Ab%5Ex first.
Then, I would substitute the coordinates of the points given and look at what I get for ideas.
0.84=ab
1.008=ab%5E2
Once you look at those two equations, you may think of a good way to solve it, that would not give you much trouble. Even if you do not choose the easiest or fastest way to the solution, do not panic. Try to stay calm and confident.
ONE OPTION
You could decide on dividing the second equation by the first one
1.008%2F0.84=ab%5E2%2Fab --> 1.008%2F0.84=b or b=1.008%2F0.84 --> b=1.2
Once you get b, you can substitute in 0.84=ab to find a.
0.84=a%2A1.2 --> 0.84%2F1.2=a or a=0.84%2F1.2 --> a=0.7
So the function is highlight%28y=0.7%2A1.2%5Ex%29
ANOTHER OPTION
You could decide on solving for a in the first equation and substituting that into the second one.
For example, you found that 0.84%2Fb=a and substituted to get
1.008=%280.84%2Fb%29%2Ab%5E2
That may look scary, but can be simplified
1.008=%280.84%2Fb%29%2Ab%5E2 --> 1.008=0.84%2Fb%2Ab%5E2 --> 1.008=0.84b%2Ab%2Fb --> 1.008=0.84b%2A%28b%2Fb%29 --> 1.008=0.84b --> b=1.008%2F0.84 b=1.2
amd you can substitute that into 0.84=ab to find a, as shown above.
This option may look more complicated, but still leads to the solution.
AND YET ANOTHER OPTION
You could decide on solving forb in the first equation and substituting that into the second one.
b=0.84%2Fa substituted into 1.008=ab%5E2
1.008=a%2A%280.84%2Fa%29%5E2 --> 1.008=a%2A%280.84%5E2%2Fa%5E2%29 --> 1.008=a%2A0.84%5E2%2Fa%5E2 --> 1.008=a%2A0.84%5E2%2F%28a%2Aa%29 --> 1.008=%28a%2Fa%29%2A%280.84%5E2%2Fa%29 --> 1.008=0.84%5E2%2Fa --> 1.008a=0.84%5E2 --> a=0.84%5E2%2F1.008 --> a=0.7
and you can substitute that into 0.84=ab to find b:
0.84=0.7b --> 0.84%2F0.7=b or b=0.84%2F0.7 --> b=1.2
COMMENT
Staying calm is difficult, but necessary. Being confident is even harder. Having seriously practiced algebra in previous years will help give you confidence, and make calculation mistakes less likely. If you did not take algebra that seriously before, I can tell you that I've been there, and that there are plenty of people in the same boat. Several times, I found out the hard way that math, like languages and music, gets much harder the longer that you do not practice it seriously and regularly. Fortunately, at various points, I stumbled into very studious study buddies that made me practice math a lot more and be (temporarily) a little more responsible.