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OpenStudy (anonymous):
@Euler271
OpenStudy (anonymous):
@Loser66
OpenStudy (loser66):
\(20Q = (1.13)^t\) ok?
OpenStudy (anonymous):
i just need to say what is the
- starting value a
- growth rate r
- & continuous growth rate k
OpenStudy (anonymous):
k then what..
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OpenStudy (loser66):
ln both sides to get t down
OpenStudy (loser66):
ln 20Q = ln (1.13)^t = t ln (1.13)
so, ???
OpenStudy (anonymous):
but its 20q so both sides have a variable.. can u still take the ln?
OpenStudy (loser66):
\(\large t = \dfrac{ln20Q}{ln(1.13}= log _{1.13}20Q\)
OpenStudy (anonymous):
ok. then?
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OpenStudy (loser66):
then waaat? I ask you what Q mean, you said "It's just a part of equation" so, I solve it as if it is a "nonsense letter" , now then??? how can I know what should I do next?
OpenStudy (anonymous):
sorry...
OpenStudy (anonymous):
well in logarithms and all that stuff they have equations like Q=ab^t or Q= ae^kt
OpenStudy (anonymous):
the general equation of this is:
\[Q = ar^{kt}\]
OpenStudy (anonymous):
ok
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OpenStudy (anonymous):
a = 1/20
r = 1.13
k = 1
OpenStudy (loser66):
oooooh. definition. got you @Euler271
OpenStudy (anonymous):
i may have inverted r and k.
yea ^_^
OpenStudy (anonymous):
k thnks @Euler271. BUT ONE QUESTION. why is k just 1?
OpenStudy (anonymous):
because of the equation you gave
since it is r^(kt)
1.13^(kt) = 1.13^(t)
k = 1
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