What was the intial temperature in the problem below?Problem:  T(x) = 60(1/2)^(x/30) + 20 , where T(x) is the temperature in degrees Celsius and x is the elapsed time in minutes. Graph the...

What was the intial temperature in the problem below?

Problem:  T(x) = 60(1/2)^(x/30) + 20 , where T(x) is the temperature in degrees Celsius and x is the elapsed time in minutes. Graph the function and determine how long it takes for the temperature to reach 28 deg Celsius.

The temperature of a cooling liquid over time can be modeled by the exponential function:

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The given equation is:

`T(x) = 60 (1/2)^(x/30) + 20`

For  T = 28 deg Celsius, x = ?

Solution:

Substitute the value of T to the above equation.

       `28 = 60 (1/2)^(x/30) + 20`

`28 - 20 = 60(1/2)^(x/30)`   

      `8 = 60(1/2)^(x/30)`  

    `8/60 = (1/2)^(x/30)`

    `2/15 = (1/2)^(x/30)`

Use the property of logarithm  `log_b m^n = n log_b m`

So, take the logarithm of both sides of the equation. For this, let's use the natural logarithm.

    `ln(2/15) = ln (1/2)^(x/30)`

    `ln(2/15) = (x/30) ln(1/2)`

     `ln(2/15) / (ln(1/2)) = x/30`

   `30 ln(2/15)/(ln(1/2)) = x`

     87.2 mins = x

 

For the initial temperature, set x=0. Then solve for T(0).

Solution:

 `T(0) = 60(1/2)^(0/30) + 20 = 60 (1/2)^0 + 20 = 60 (1) + 20 = 80 deg Celsius`

Answer: The intial temperature is 80 deg Celsius. To lower the temperature from 80 to 28 deg Celsisus, it takes 87.2 minutes.

 

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