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Mathematics 18 Online
OpenStudy (anonymous):

Q = ((1.13)^t) / 20

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..

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?

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

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

OpenStudy (anonymous):

i thought itd be ln1.13

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