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
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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):
\[70 = ab^{10}\]\[10 = ab^{30}\]
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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
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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?
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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*
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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
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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) ?
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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)
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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)
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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
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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
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OpenStudy (anonymous):
(/20)
OpenStudy (amistre64):
(1/20) i was wanting to do multiply by x and make it x/20 but my fingers protested