# What is the value of $a$ & $r$ when 2 points on the graph are given? [closed]

I am working on algebraic functions and I am stuck on this problem:

$f(x) = a * r^x$
$(2,1),(3,1.5)$

This would be a simple problem if it weren't for that $1.5$ -

So, I have plugged in $2$ and $1$ into the function and this is what I got:

[I also solved for $3,1.5$]
So, this is where I get stuck every time.

I have tried solving it multiple ways but it always ends up to the same effect, where $r$ is always $=$ to something funny that is unsolvable - at least for me.

Could someone explain how you got your answer so that I can understand the process? I know how to do this type of problem, but it is just this one that I can't get!!!!

Thanks for your help and time!

EDIT: I need this simplified, that is why I cant leave $r^2 = 1.5$

## closed as unclear what you're asking by Alex M., user99914, Daniel W. Farlow, Claude Leibovici, user91500Aug 11 '16 at 5:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Just for community advice: Drawings are typically taboo in Math SE mainly because there are mobile users that use Math SE and they won't be able to see any pictures that you post in your question. While I'm not saying that you should get rid of your picture, I would just like for you to be mindful of this. – KingDuken Aug 10 '16 at 14:27
• @KingDuken - thank you for the information. – Carlos Carlsen Aug 11 '16 at 15:17

So if we have the function f(x)=ar^x, then by substituting the first point given, namely $(2,1)$, we'll get:

$$1=ar^2$$

Similarly, substituting $(3, \frac{3}{2})$, we end up with

$$\frac{3}{2}=ar^3$$

Now, to find $a\neq 0$ and $r\neq 0$, you should divide this two equations, what you get then?

• $r = 3/2$ $\Rightarrow$ we can find the $a$!!! – Carlos Carlsen Aug 10 '16 at 14:30
• But why couldn't I find the square root of 1 and get $ar = 1$? Or is that not the proper way to do it? – Carlos Carlsen Aug 10 '16 at 14:32
• I think that the idea of your approach was correct, but something went wrong implementing that, i.e: you want to write $a=\frac{1}{r^2}$, and the substitute it to the second equation, right? So, $\frac{3}{2}=\frac{1}{r^2}\cdot r^3$, thus $\frac{3}{2}=r$. – Salech Rubenstein Aug 10 '16 at 14:38
• So the way I solved for $r$ in $ar^2 = 1$ was incorrect, I assume? – Carlos Carlsen Aug 10 '16 at 14:50

$$f(2)=ar^2=1$$ $$a=\frac{1}{r^2}$$ $$f(3)=ar^{3}=1.5$$ $$r=1.5$$ $$a=\frac{1}{(1.5)^2}$$

• Thank you for your help, though I don't understand one part of you solution. How did you get $r = 1.5$ from $ar^3 = 1.5$? – Carlos Carlsen Aug 10 '16 at 14:48
• @CarlosCarlsen Because $a=\frac{1}{r^2}$. On substituting, you get $r=1.5$. – GoodDeeds Aug 10 '16 at 15:05