A cup of tea has a temperature of 100 degrees Celsius. it is cooling in a room that has a temperature of 20 degrees Celsius. it cools in accordance with the following model..

T(t) = 100 - 70 (1- (1/2) ^ t/3 )

T = temp in degrees Celsius

t = time in minutes

in its simplified form `T(t) = 70(1/2) ^(t/3) + 30`

determin when T= 50 degrees Celsius

### 1 Answer | Add Yours

`T(t)=70(1/2)^(t/3)+30`

`` Determine t = ? when T=50 degree Celsius.

`50=70(1/2)^(t/3)+30`

`50-30=70(1/2)^(t/3)`

`20=70(1/2)^(t/3)`

`20/70=(1/2)^(t/3)`

`(2/7)^3=(1/2)^t`

`.0233=(1/2)^t`

`log(.0233)=tlog(1/2)=t(log(1)-log(2))`

`log(.0233)=-t log(2)`

`t=log(.0233)/(-log(2))`

`t=5.42` min.

Notice

log(1)=0

and

if

`x^(1/n)=y`

`x=y^n`

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