Question

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.

Answer 1

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

Answer 2

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

Answer 3

thanks alot damon ...i luv u!!!!