Question

does anyone know how to find the formula for exponential functions that satisfy these given conditions?
f(0)=4; F(1)=2
h(0)=100; h(1)=75
p(1)=4; p(2)=6

Answer 1

(0,4) (1,2)

we know that an exponent is normally at (0,1) so this has been modified at least by +3

Answer 2

(0,4-3) (1,2-3) falls below the range tho :)

we can try using the points to figure it out :) wrap an exponent to fit

Answer 3

or find the inverse; with logs :)

Answer 4

f(4) = 0 ; f(2) = 1

logb(4) = 0 ; logb(2) = 1

Answer 5

b^1 = 2 suggests that b=2...but its modified :) so that cant be right.

b^0 = 4 suggests that we are b^x + 3

y = b^x +3

y-3 = b^x

2-3 = b^(1)

b = -1 ??

Answer 6

y = -(1)^x +3 is what I seem to get :)

Answer 7

An exponential function has the form:
\[y=a.b^x\]
You just need to substitute using the two points given in each case, and solve for a, b.
I will do the first one in the next post.

Answer 8

yay!! someone smarter than me :)

Answer 9

im guessing a = 4 ;)

Answer 10

For the first one, we have the following two equations:
\[4=a.b^0 \rightarrow(1)\]
and
\[2=a.b^1\rightarrow(2)\]

Answer 11

So clearly, as amistre64 just said, since b^0=1, we have a=4. Then for equation (2):
\[2=4b \implies b={2 \over 4}= {1 \over 2}\]

Answer 12

So we have a=4 and b=1/2, then the exponential function is:
\[f(x)=4({1 \over 2})^x\]

Answer 13

Does that make sense to you? and @amistre64: I am not smarter than you :)

Answer 14

The answer does... I'm not sure I understand how you got it though

Answer 15

Which step don't you understand?

Answer 16

You first should know that the form of an exponential function is:
\[f(x)=a.b^x\]
where a and b are constants.

Answer 17

well for the next one, h would also equil to ^0 so 1 so does that make it F(x)=100x 75^1

Answer 18

So, your goal from here is to find these two constants using the two given points in each case.

Answer 19

?

Answer 20

okay

Answer 21

The formula makes sence to me, its just the way the points fit into it that confuses me

Answer 22

I will tell you how to substitute the points. Let's take the point f(0)=4, that says when you plug x=0, you will get value f(x), h(x) (or whatever you call it) to be 4.
So the number between the brackets goes for the exponent.

Answer 23

So for h(0)= 100... H=1 because anything raised to the zero power is one then times 100 which is 100? =/

Answer 24

Not quite, you should always think of the formula. Now, the formula looks like:
\[h(x)=a.b^x\]
Try to use the points to find a and b.

Answer 25

Let me show you how to use the first point to find a, and then you should try finding b.
The point is h(0)=100. So, plug x=0, and 100 for h(x).

Answer 26

100 x 1?

Answer 27

So, doing what I just said. We have h(x)=ab^x where x=0, and h(x)=100. that gives:
\[h(x)=a.b^x \implies 100=a.b^0\]
but b^0=1, then
\[100=a(1) \implies a=100\]

Answer 28

We just did about 75% of the problem. Don't forget that you're looking for values for a and b, and then plug them back in the formula. That's all. Can you use the second point to find b?

Answer 29

would it be 75^1=75?... that seems too simple.

Answer 30

Don't forget you want to find b, so try to plug in the formula and tell me what b would be.

Answer 31

Take a breath :)

Answer 32

Look closely to the formula:
\[h(x)=a.b^x\]

Answer 33

100 x (75)^1?

Answer 34

We already found a=100, so we can write it as:
\[h(x)=100(b)^x\] RIGHT?

Answer 35

yes

Answer 36

So now we will use the second point h(1)=75 to find b. We should put x=1 and h(x)=75.
You're close by the way :)

Answer 37

75/100?

Answer 38

Exactly.
\[75=100(b)^1 \implies 75=100b \implies b={75 \over 100}\]
But you can simplify this value a little bit of you like :)

Answer 39

Simplified b=(3/4).

Answer 40

So, the equation would look like?!

Answer 41

I hope I didn't complicate things to you. Anyway, all you have to do now is to substitute these to values a=100, b=3/4 into our original function, you get:
\[h(x)=100({3 \over 4})^x \]