Question

How do you find the exponential model `y=ae^(bx)` that goes through the points (0,7) (5, 32)?

Answer 1

`y = 7e^(1/5ln(32/7)x)`

Explanation:

We can solve for `a` without writing a systems of equations. Notice in the first point, `x = 0`. This will make `b = 0` since `0 xx a = 0`. We will be left with only `a`.

`y = ae^(bx)`

`7 = ae^(b xx 0)`

`7 = ae^0`

`7 = a(1)`

`a = 7`

Substituting, and solving for b:

`y = ae^(bx)`

`32 = 7e^(5b)`

`32/7 = e^(5b)`

`ln(32/7) = ln(e^(5b))`

`ln(32/7) = 5b(lne)`

`ln(32/7) = 5b`

`(ln(32/7))/5 = b`

`1/5ln(32/7) = b`

Hence, the equation of the function is

`y = 7e^(1/5ln(32/7)x)`

Hopefully this helps!

Answer 2

`y=7(2*7^(-1/5))^x`.

Explanation:

Let us denote, by `C`, the curve represented by `: y=ae^(bx)...(0)`.

The pt. `(0,7) in C rArr "its co-ords. must satisfy (0)"`.

`rArr 7=ae^0=a.........(1)`.

`" pt."(5,32) in C rArr 32=ae^(5b)`.

By `(1) "then", e^(5b)=32/7, or, e^b=(32/7)^(1/5)`.

Hence, `(0)` becomes,

`y=ae^(bx)=7(e^b)^x=7{(32/7)^(1/5)}^x=7(2*7^(-1/5))^x`.