1) given that x= log(small5)3 + log(small5)4, find algebraically the value of x
2)a mug of tea cools according to the law T(small t)= T(small 0)e to the power^-kt where T(small 0) is the initial temp and T(small c)is the temp after t minutes. all temps are in celciius.
a) a particular mug of tea is cooled from 100*C to 75*C in a quater of an hour. therefore clculate value of k.
b) by how many degrees will the temperature of this tea fall in the next quater of an hour.
you want log5(12)
x = log5 (12)
or
5^x = 12
log10 (5^x) = log10 (12) = 1.0792
or
x log10 (5) = log10 (12) = 1.0792
x (.699) = 1.0792
x = 1.0792/.699
2)a mug of tea cools according to the law T(small t)= T(small 0)e to the power^-kt where T(small 0) is the initial temp and T(small c)is the temp after t minutes. all temps are in celciius.
Tt = To e^(-kt)
a) a particular mug of tea is cooled from 100*C to 75*C in a quater of an hour. therefore clculate value of k.
75 = 100 e^-(.25k) if t in hours and not minutes
.75 = e^-.25k
ln .75 = -.25k
k = -.288/-.25
b) by how many degrees will the temperature of this tea fall in the next quater of an hour.
T = 75 e^-1.15 (.25)
I will leave that to you
thanks alot damon ...i luv u!!!!
1) To find the value of x, we'll start by simplifying the given expression using the properties of logarithms. The property we'll use is the addition property:
log(a) b + log(a) c = log(a) (b * c)
Using this property, we can combine the two logarithms in the expression:
x = log(small5)3 + log(small5)4
x = log(small5) (3 * 4)
x = log(small5) 12
Now we need to evaluate the logarithm base small5 of 12. Since x is an exponent that we need to raise small5 to in order to get 12, we can rewrite the equation as:
5^x = 12
To solve for x, we can take the logarithm of both sides with the base 5:
log(small5) (5^x) = log(small5) 12
x = log(small5) 12
So the algebraic value of x is log(small5) 12.
2a) The cooling of the tea follows the equation T(small t) = T(small 0) * e^(-kt), where T(small t) is the temperature after t minutes, T(small 0) is the initial temperature, and k is the rate at which the tea cools.
We're given that the tea is cooled from 100°C to 75°C in a quarter of an hour, which is 15 minutes. Using this information, let's set up the equation:
75 = 100 * e^(-k * 15)
Now we can solve for k by rearranging the equation:
e^(-k * 15) = 75/100
e^(-k * 15) = 0.75
To isolate k, we can take the natural logarithm (ln) of both sides:
ln(e^(-k * 15)) = ln(0.75)
-k * 15 = ln(0.75)
k = -(1/15) * ln(0.75)
So the value of k is -(1/15) * ln(0.75).
2b) To calculate the number of degrees the temperature of the tea will fall in the next quarter of an hour, we need to find T(small 15) (the temperature after 15 minutes) and T(small 30) (the temperature after 30 minutes).
Using the given equation T(small t) = T(small 0) * e^(-kt), and the value of k we calculated in part 2a, we can substitute t = 15 and t = 30:
T(small 15) = T(small 0) * e^(-(1/15) * ln(0.75))
T(small 30) = T(small 0) * e^(-(1/15) * ln(0.75)) * e^(-(1/15) * ln(0.75))
To calculate the difference in temperature between T(small 30) and T(small 15), subtract T(small 30) from T(small 15):
Temperature difference = T(small 15) - T(small 30)
Calculating this difference will give you the number of degrees the temperature of the tea will fall in the next quarter of an hour.