A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
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
anonymous
 5 years ago
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

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(0,4) (1,2) we know that an exponent is normally at (0,1) so this has been modified at least by +3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(0,43) (1,23) falls below the range tho :) we can try using the points to figure it out :) wrap an exponent to fit

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or find the inverse; with logs :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0f(4) = 0 ; f(2) = 1 logb(4) = 0 ; logb(2) = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0b^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 y3 = b^x 23 = b^(1) b = 1 ??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = (1)^x +3 is what I seem to get :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0An 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.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yay!! someone smarter than me :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For the first one, we have the following two equations: \[4=a.b^0 \rightarrow(1)\] and \[2=a.b^1\rightarrow(2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So clearly, as amistre64 just said, since b^0=1, we have a=4. Then for equation (2): \[2=4b \implies b={2 \over 4}= {1 \over 2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we have a=4 and b=1/2, then the exponential function is: \[f(x)=4({1 \over 2})^x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that make sense to you? and @amistre64: I am not smarter than you :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The answer does... I'm not sure I understand how you got it though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which step don't you understand?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You first should know that the form of an exponential function is: \[f(x)=a.b^x\] where a and b are constants.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well for the next one, h would also equil to ^0 so 1 so does that make it F(x)=100x 75^1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, your goal from here is to find these two constants using the two given points in each case.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The formula makes sence to me, its just the way the points fit into it that confuses me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So for h(0)= 100... H=1 because anything raised to the zero power is one then times 100 which is 100? =/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Not 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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 \[100=a(1) \implies a=100\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would it be 75^1=75?... that seems too simple.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't forget you want to find b, so try to plug in the formula and tell me what b would be.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Look closely to the formula: \[h(x)=a.b^x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We already found a=100, so we can write it as: \[h(x)=100(b)^x\] RIGHT?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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 :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly. \[75=100(b)^1 \implies 75=100b \implies b={75 \over 100}\] But you can simplify this value a little bit of you like :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, the equation would look like?!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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: \[h(x)=100({3 \over 4})^x \]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.