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/sci/ - Science & Math

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2593154 No.2593154 [Reply] [Original]

Find a formula for f(x), an exponential function such that f(3)=4 and f(5)=64.

How smart are you /sci/? Think you can do something as simple as precal?

>> No.2593197

>>2593194
Show your work, son.

>> No.2593194

lol i think i fucked it up:

0.0625(4)^x

>> No.2593211

>>2593197
(ab^5)/(ab^3)=64/4
b^2=16
b=4
a(4)^3=4
a=4/(4^3)
a=0.0625

/work

>> No.2593223

Boom. That's the answer, plug in the x value to check your work. Good job anon.

>> No.2593251

Anybody care to guide me through the solution progress step by step? For instance, how do you get the exponentials 5 and 3?

>> No.2593252
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2593252

>>2593197

Homework thread eh?

>> No.2593322

>>2593251
Bumping for help on how to solve this.

>> No.2593445

>>2593322
Shameless bump again.

>> No.2593456

>>2593251
they're the x values bro. the equation for an exponential function is ab^x. You just have to plug everything in. Just look at >>2593211 and you'll see how he figured it out

>> No.2593467

>>2593456
Yeah, I'm wondering where he got 5 and 3 from.

>>(ab^5)/(ab^3)=64/4

This is the part which eludes me.

>> No.2593470

f(x) = 2^(x-1)
that was easy

>> No.2593549

>>2593470
16 != 64.

>> No.2593590

f(x)=ab^x
f(3) = 4 --> ab^3 = 4
f(5) = 64 --> ab^5 = 64
(b^5)/(b^3)=64/4
b^2 = 16 --> b = 4

a(4)^3 = 4
a = 1/16

f(x)=(1/16)(4^x) OR f(x)=4^(x-2)
Yup.

>> No.2593707

>>2593470
>>2593470

pro skills. f(x) = 2^(n-1)?

f(5) = 2^(5-1)
= 2^4
= 16
>>f(5)=64.
derp.

>> No.2593719

f(x) = 4^(n-2)

>> No.2593726

>>2593470
moar like f(x) = 2^(2x-4)
lrn2powersoftwo

>> No.2593750

>>2593194
he has the right equation. everyone else messed up lol

>> No.2593762

>>2593750
it works, but so does mine >>2593719

>> No.2593785
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2593785

>>2593750
>mfw only one other person messed up and all other answers are equivalent forms

>> No.2593778

>>2593467
read the problem again?
f(3)=4 and f(5)=64
he got the 3 and 5 from that
not sure how you don't see that?
3 and 5 are the x values and since the equation is ab^x you plug the 3 and 5 into the equation, making it ab^5/ab^3, then you equal that to the y value, 64/4 so in the end it all looks like this:
ab^5/ab^3=64/4
b^2 because you subtract exponents and 64/4=18 so
b^2=18
b=18^(1/2) you do this to get rid of the exponent
b=4

get it?

>> No.2593780
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2593780

infinitely many answers anons

troll

>> No.2593792

>>2593778
lol only helpful nontrolling anon

>> No.2593797

>>2593792
i'm a helpful nontrolling tripfag...

its fucking f(x) = 4^(x-2)

>> No.2593808

Legitimate question:
How do we know this is an exponential function?

>> No.2593810

>>2593797
show your work son

>> No.2593814

>>2593808
because it is stated in the OP.
it is one of the condistions.

>> No.2593818

>>2593797
i love how on /sci/ if you come up with the best answer you mostly get ignored

>> No.2593823

>>2593810
iunno..just logic.
>>2593780

try inputting x = 3 or 5 into it, and you can see that it works

...altho probably infinite answers, because this.

>> No.2593824

>>2593814
Oops, I just read the equation.
I'm bad for only seeking out numbers in math questions.

>> No.2593825

>>2593808
Did you read the question you fuckface?
>Find a formula for f(x), an exponential function such that f(3)=4 and f(5)=64
>an exponential function such that f(3)=4 and f(5)=64
>an exponential function
>exponential function
>exponential

>> No.2593829
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2593829

>>2593818
i'm used to getting ignored...

>> No.2593837

>>2593823
i was ignoring you because you didn't show your work. even if it is logic, it's still nice to have evidence (excluding inputing the numbers into the equation)

>> No.2593839

ITT: attention whoring tripfags
Also, the correct answer given and thoroughly explained in six different places.

>> No.2593844

>>2593839
ITT: people that have so much time on their hands that they count how many times the answer was given to a simple problem on a forum post no one cares about

>> No.2593846

>>2593829
i'm guessing that you got it by inspection as i did.

not using the rote learnt method all the tards are using, which gives the equivalent answer of 1/16 4^x .

there is only 1 answer though, just lots of equivalent forms of it.

>> No.2593855

given that the equation is of the form y=a*b^(x-c), there are infinite solutions because you can change either a or c to be anything and then calculate the other. because of this, it's usually best to stick to the form y=a*b^x.

>> No.2593857

f(3)=4, f(5)=64, f(x)=Ae^kx

4=Ae^3k
64=Ae^5k

64/4=Ae^5k / Ae^3k
16=e^2k
ln16 /2=k

f(x)=Ae^ln16x /2
f(x)=A(4)^x
f(3)=4
4=A(4)^3
4=64A
A=1/16
f(x)=1/16 (4)^x
bitches don't know about my proper form

>> No.2593867

>>2593855
>>2593855
fullretard.jpg

>> No.2593874
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2593874

>>2593837
question is simple enough that showing work is not necessary.

>>2593839

well im the only tripfag here, and i am not attention whoring, i am just correct,
fuck off.

>>2593846
correct. :)

>> No.2593875

>>2593855
c gets absorbed by a, herpa derpa

all forms are equivalent

>> No.2593911

4^(x-2)

laaaash