g(x) = 23(.94)^x give the temperture in degrees Celsius of a bowl of water x minutes after a large quantity of ice is added. After how many minutes will the water reach 5 degrees Celsius?
2007-02-06 03:20:00 · 6 answers · asked by *only~wishful~thinking* 3 in Science & Mathematics ➔ Mathematics
5=23(.94)^x
5/23=.94^x
lg(5/23)=xlg.94
x=lg(5/23)/lg(.94)
2007-02-06 03:26:04 · answer #1 · answered by Maths Rocks 4 · 0⤊ 0⤋
g(x) is the temperature.
It is given g = 5 when the time is x minutes.
5 = 23 * (.94) ^x.
Taking logarithms to the base 10 on both sides.
log 5 = log 23 + log (.94) ^x
log 5 - log 23 = log (.94) ^x
log 5 - log 23 = x*log (.94)
x*log (.94) = log 5 - log 23
x*(-0.0269) = - 0.6628
x = 24 .67 minutes.
2007-02-06 04:40:37 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
with this formula the temperature at x=0 would be 23 oC
5 = 23*(0.94)^x Taking log of both sides
log(5/23)= x log 0.94 so x= log(5/23)/log 0.94 =24.66minutes
2007-02-06 03:48:54 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
Temperature: g(x) = 23 * (0.94)^x
For g(x) = 5,
23 * (0.94)^x = 5
(0.94)^x = 5/23
Taking log at both sides,
log(0.94^x) = log(5/23)
x * log(0.94) = log(5/23)
x = log(5/23) / log(0.94)
Use a scientific calculator to evaluate the logs above and get the final result, in minutes. Don't forget to convert the decimal part to seconds.
2007-02-06 03:39:38 · answer #4 · answered by jcastro 6 · 0⤊ 0⤋
5= 23(.94)^x
5/23 = (.94)^x
log(base .94) 5/23 = x
x = log (5/23) / log (.94)
where log is log (base 10)
2007-02-06 03:25:38 · answer #5 · answered by bequalming 5 · 1⤊ 0⤋
24.66347227 minutes.
2007-02-06 03:25:56 · answer #6 · answered by UnENG 3 · 1⤊ 0⤋