Question 1069541: 1. log x - log (x+2)=1
2. (4^x)-40=3*2^x
3.Det. Robert Goren is investigating a homicide, and the medical examiner tells him that when he arrived at 6:00 p.m., the body had a temperature of 95.3 degrees Fahrenheit. By the time Nichols had arrived an hour later (at 7:00), he checks again and determines the temperature to be 94.9 degrees. The ambient air temperature is 90 degrees. Use the equation 𝑇(𝑡) = 𝑎𝑒^(−𝑘𝑡) + 90 to determine when the victim was killed. (The normal body temperature is 98.6 degrees. Round k to 3 decimal places and find the time of death to the nearest minute.
For number 1, I got -20/9, but it can't be negative so I'm confused. Thank you for answering!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 1. log x - log (x+2)=1
Assuming base 10:
log(x) - log(x+2)=1 = log(10)
log(x/(x+2)) = log(10)
x/(x+2) = 10
x = 10x + 20
x = -20/9
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But, log(-20/9) is not allowed.
--> no solution.
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2. (4^x)-40=3*2^x
2^(2x)-40 = 3*2^x
2^(2x) - 3*2^x - 40 = 0
(2^x - 8)*(2^x + 5) = 0
2^x = 8
x = 3
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2^x = -5 ignore
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3.Det. Robert Goren is investigating a homicide, and the medical examiner tells him that when he arrived at 6:00 p.m., the body had a temperature of 95.3 degrees Fahrenheit. An hour later (at 7:00), he checks again and determines the temperature to be 94.9 degrees. The ambient air temperature is 90 degrees. Use the equation D(t) = a*e^(-kt) + 90 to determine when the victim was killed. (The normal body temperature is 98.6 degrees. Round k to 3 decimal places and find the time of death to the nearest minute.
I'll use D for degrees.
D(t) = a*e^(-kt) + 90
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D(0) = a*e^(-kt) + 90 = 95.3 (At 6PM)
a*e^(-kt) + 90 = 95.3
Have to finish later.
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For number 1, I got -20/9, but it can't be negative so I'm confused.
The solution is negative --> the equation is not valid.
--> no solution.
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