Question

What is the exponential function in the form `f(x)=ab^x` and has values: `f(1)=2` and `f(2)=5`?

Answer 1

`f(x) = (4/5)(5/2)^x`

Explanation:

Given: `f(x)=ab^x`

The point `f(1) = 2` yields equation [1]:

`2 = ab^1" [1]"`

The point `f(2)=5` yields equation [2]:

`5 = ab^2" [2]"`

We can temporarily eliminate the variable "a" and find the value of "b" by dividing equation [2] by equation [1]:

`5/2= (ab^2)/(ab^1)`

`b = 5/2`

Substitute 5/2 for "b" in equation [1] and then solve for "a":

`2 = a(5/2)`

`a = 4/5`

Substitute the values for "a" and "b" into `f(x)`:

`f(x) = (4/5)(5/2)^x`

Here is the graph:

graph{(4/5)(5/2)^x [-10, 10, -6, 6]}

Answer 2

The answer is `f(x)=(4/5)(5/2)^x`

Explanation:

Express the function you are trying to find as `y = ab^x`
Substitute `(1,2)` in the equation we just expressed.
That is, where there is y, put in 2 and where there is x put in 1.
Now, we have `2 = ab` (let this be equation 1)

Then, substitute `(2,5)` in the equation we expressed in.
That is, put 2 where we have x and 5 where we have y:
`5=ab^2` (let this be equation 2).

Now you need to solve these equations simultaneously. You could use the substitution method, but the most efficient method in this case is elimination. You can tell which method to use by assessing the equations you are given - use elimination if both equations are linear or if one term is eliminated one term when dividing the equation by the other. Substitution is generally used when you have several terms of various degrees.

Now then, divide equation 2 by equation 1:
`5/2 = (ab^2)/(ab)`
using exponential laws, `b = 5/2`

This means that you have `y = a(5/2)^x`

Now take either of the points and plug it into the newly found equation. However, it is simpler, quicker and more efficient to take `(1,2)` in this case. So, we have:
`2=a(5/2)`
Therefore, `a=4/5`

Now we take the values of `a` and `b` and plug them in `f`. That is, `f(x)=(4/5)(5/2)^x`

You can check if this is true by taking one of the given x-values and plugging them in where there is x - you will get the corresponding y-value.