Bunuel wrote:
kilukilam wrote:
The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60 , where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite
a. 65
b. 75
c. 80
d. 85
e. 90
We have the formula for calculating temperature: \(F=120*2^{-at}+60\) and the value for \(t=10\) --> \(F(t=10)=120=120*2^{-10a}+60\) --> \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\) --> \(a=0.1\). So formula is \(F=120*2^{-0.1t}+60\)
Now for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).
Answer: B.
You're solution is quite clear to me, I only have one question:
How do you conduct this step: \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\)
I would solve it like this:
\(\frac{1}{2^{10a}}=\frac{1}{2}\) --> 2^-1 = 2^-10a --> -1 lg(2) = -10a lg(2) --> 1 = 10aIs there a easier possibility? Is it even right to solve the question by the use of logarithm?
Thanks in advance!