Question 40754: I desperately need help with these few questions. This is a take home test that needs to be done by tommorow.
1. The number of bacteria present in a culture is B=10e^1923t where t is the time in minutes. Find the time required, to the nearest half minute, to have 3200 bacteria present.
12. log(5x-1)=2+log(x-2)
13. log(x^3)=(logx)^2
Thank you Very Much,
Art
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. The number of bacteria present in a culture is B=10e^1923t where t is the time in minutes. Find the time required, to the nearest half minute, to have 3200 bacteria present.
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3200=10e^(1932t)
320=e^(1932t)
Take the natural log of both sides to get:
ln(320)=1932t
t=5.76832.../1932=0.00298567...
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12. log(5x-1)=2+log(x-2)
log(5x-1)-log(x-2)=2
log[(5x-1)/(x-2)]=2
(5x-1)/(x-2)=10^2
5x-1=100(x-2)
5x-1-100x+200=0
-95x=-199
x=2.0947...
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13. log(x^3)=(logx)^2
3logx =(logx)^2
(logx)^2-3logx=0
Factor out the logx to get:
(logx)(logx-3)=0
logx=0 or logx=3
If logx=0 10^0=x. Therefore x=1
If logx=3 10^3=x. Therefore x=1000
Cheers,
Stan H.
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