Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):

Please help, im going to cry. Find a formula for the exponential function which satisfies the given conditions: g(10)=70 and g(30)=10

OpenStudy (anonymous):

I have no clue what im doing

OpenStudy (paxpolaris):

do you know what the general form of an exponential function looks like.

OpenStudy (anonymous):

yes f(x) = a(b)^x

OpenStudy (anonymous):

So f(x) is 70 and 10, the x is the 10 and 30

OpenStudy (paxpolaris):

\[70 = ab^{10}\]\[10 = ab^{30}\]

OpenStudy (amistre64):

a b^(x-h) + k is what i was thinking

OpenStudy (anonymous):

thats vertex form?

OpenStudy (anonymous):

nvm lol

OpenStudy (amistre64):

dunno the name of it, it just covers all the possible shifts and scalars

OpenStudy (paxpolaris):

we don't have enuf info to use your formula @amistre64

OpenStudy (amistre64):

bummer .... im not opposed to h,k = 0,0

OpenStudy (paxpolaris):

@TheRealBMH .. so take the log of both side of both equation and solve.

OpenStudy (anonymous):

what does the log of ab^2 look like?

OpenStudy (paxpolaris):

loga +2 logb

OpenStudy (anonymous):

1.85 = loga + 2 logb?

OpenStudy (paxpolaris):

where you get 2 from though????

OpenStudy (amistre64):

we could sub it as well, then log out of it maybe 70/b^10 = a 10 = 70b^(20)

OpenStudy (anonymous):

from you lol

OpenStudy (anonymous):

ohhh xD

OpenStudy (anonymous):

x*

OpenStudy (paxpolaris):

substituting first seems better :)

OpenStudy (anonymous):

where di the 20 come from?

OpenStudy (amistre64):

b^(30) ------ b^(10)

OpenStudy (amistre64):

solving for a, then subing it in

OpenStudy (amistre64):

30 bs, canceled by 10 bs leaves 20 bs

OpenStudy (anonymous):

ok so a is 20 now we just solve for b?

OpenStudy (amistre64):

70 = a b^10 , solve for a by dividing off b^10 a = 70/b^10 sub this into the other setup 10 = a b^30 10 = 70 b^20 1/7 = b^20, not take the 20th rt b = (1/7)^(1/20)

OpenStudy (amistre64):

*now take ....

OpenStudy (amistre64):

is spose since 20 is even, it should be +- 1/7 but i never worryabout stuff like that

OpenStudy (anonymous):

so 20(1/7)^(1/20) ?

OpenStudy (anonymous):

i feel so dumb

OpenStudy (xapproachesinfinity):

1/7 = b^20, not take the 20th rt you have to be careful here?

OpenStudy (xapproachesinfinity):

the sign of b

OpenStudy (anonymous):

so i just leave b how it is??

OpenStudy (amistre64):

well, we know a value for b, and since b gets smaller as x gets bigger, lets go negative and a = 70/b^10 a = -70/(1/7)^(1/20)

OpenStudy (anonymous):

=-63.5

OpenStudy (amistre64):

forgot the ^10 a = -70/(1/7)^(1/2)

OpenStudy (anonymous):

-185.2

OpenStudy (anonymous):

and b was 20?

OpenStudy (amistre64):

1/7 = b^(20)

OpenStudy (anonymous):

but how do i get it to f(x) = ab^2

OpenStudy (amistre64):

why do you need it as ^2 ?

OpenStudy (amistre64):

ab^2 is not an exponential function of x

OpenStudy (amistre64):

a b^x is

OpenStudy (anonymous):

ab^x

OpenStudy (anonymous):

sorry idk why i keep saying that

OpenStudy (amistre64):

we are solving for a and b

OpenStudy (anonymous):

right

OpenStudy (amistre64):

a = 70sqrt(7) b = +- (1/7)^(1/20) b^x = +- [(1/7)^(/20)]^x

OpenStudy (anonymous):

is that meant to be 20 or .20 on the last part

OpenStudy (anonymous):

(/20)

OpenStudy (amistre64):

(1/20) i was wanting to do multiply by x and make it x/20 but my fingers protested

OpenStudy (paxpolaris):

b has to be positive in an exponential function... no +/-

OpenStudy (anonymous):

exactly, thank you for pointing that out, i appreciate your help dude

OpenStudy (anonymous):

thanks for being patient and really helping and explaining @amistre64 i really appreciate it!

OpenStudy (amistre64):

youre welcome, im glad the site didnt mess up to bad along the way

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!