Curve Fitting project exponential variable equations

[alekthelion]

I am suppose to plug in two (x,y) points from the graph into the equation and solve for A first: y =Ab^x . My points are (5,4677) and (3,1527).
My first equation is 4677= Ab^5
The second is 1527= Ab^3
I am supposed to solve by substitution with the hint that I should solve for A then use the definition to find b.
Any help would be appreciated.
-Thanks

 

[Mrspi]

I am suppose to plug in two (x,y) points from the graph into the equation and solve for A first: y =Ab^x . My points are (5,4677) and (3,1527).
My first equation is 4677= Ab^5
The second is 1527= Ab^3
I am supposed to solve by substitution with the hint that I should solve for A then use the definition to find b.
Any help would be appreciated.
-Thanks

That is a FINE hint! Did you do what the hint suggested?

I'll see if I can get you started. Let's take the second equation, and solve it for A:

1527 = Ab³
Divide both sides by b³:

1527/b³ = (A*b³) / b³

And we have

A = 1527 / b³

It is solved for A (and that wasn't too hard, was it?)

The second part of the hint says to "substitute for A in the other equation." So you need to substitute 1527/b³ for A in the other equation....

4677 = A*b⁵

4677 = (1527/b³)*b⁵

Ok....you finish it!

 

[alekthelion]

So I go 4677= 1527/b^3 *b^5

4677= 1527b^5/b^3

4677= 1527b^2

4677/1527= b^2

3.06~b^2

1.75~ b?