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OpenStudy (anonymous):
wats ur qstn ?
OpenStudy (lgbasallote):
we're here for you. aal izz well
OpenStudy (anonymous):
that is why I am here for
OpenStudy (anonymous):
describe how to find the multiplier for an exponential growth rate of 15%
OpenStudy (lgbasallote):
do you mean the growth constant?
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OpenStudy (anonymous):
15/100
OpenStudy (anonymous):
0.15
OpenStudy (anonymous):
my mistake
OpenStudy (lgbasallote):
@chikitagee what's the whole problem? is the time given? we cant solve it without time
OpenStudy (anonymous):
15/100 +1 = 1.15
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OpenStudy (anonymous):
it just says to describe how to find the multiplier for an exponential growth rate of 15% and describe how to find the multiplier for a rate of decay of 15%
OpenStudy (anonymous):
for growth, 15/100+1 = 1.15
for decay, 1- 15/100 =0.85
OpenStudy (lgbasallote):
oh wait i dont think time is needed. it's already implicity given
when time = 1 then the population would be x = x_0 + 0.15 x_0
where x_0 is the initial population
now use the formula
\[\ln (\frac{x_0}{x}) = kt\]
like i said x = x_0 + 0.15 x_0 and t = 1
\[\ln (\frac{x_0}{x_0 - 0.15 x_0}) = k\]
\[\ln (\frac{\cancel{x_0}}{\cancel{x_0}(1 - 0.15)}) = k\]
\[k = \ln (\frac{1}{1- 0.15})\]
does that help?
mathslover (mathslover):
I think lgba has got the best soln .. gr8 lgba
@chikitagee got it ?
OpenStudy (anonymous):
yea it does thanks everyone
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OpenStudy (lgbasallote):
<tips imaginary hat>
OpenStudy (lgbasallote):
coincidentally...i have to leave now and get home so bye!