Math  /  Algebra

QuestionAs a hot bowl of soup is allowed to cool, its temperature TT (in degrees Fahrenheit) after tt minutes is given by the function T(t)=65+145e0.05tT(t)=65+145 e^{-0.05 t}. How long does it take for the soup to cool to 100F100^{\circ} \mathrm{F} ?
The decibel scale for measuring the intensity of sound is a logarithmic scale defined by the formula β=\beta= 10log(11012)10 \log \left(\frac{1}{10^{-12}}\right), where β\beta is the intensity of the sound in decibels (dB)(\mathrm{dB}) and // is the intensity of the sound in watts per square meter (W/m2)\left(\mathrm{W} / \mathrm{m}^{2}\right). If the sound of heavy traffic is measured to be 80 dB , what is the sound intensity in W/m2\mathrm{W} / \mathrm{m}^{2} ?

Studdy Solution

STEP 1

What is this asking? We're given a formula for how the temperature of some soup changes over time, and we need to figure out how many minutes it takes to reach 100 degrees Fahrenheit.
We're also given a formula about sound intensity in decibels, and we need to find the sound intensity in watts per square meter if the sound level is 80 dB. Watch out! Don't forget to keep track of units (minutes, degrees Fahrenheit, decibels, and watts per square meter) and make sure your final answer makes sense in the context of the problem!

STEP 2

1. Soup Cooling Time
2. Sound Intensity

STEP 3

Alright, so we're given this fancy formula for the soup's temperature: T(t)=65+145e0.05tT(t) = 65 + 145e^{-0.05t}.
Remember, T(t)T(t) is the temperature after *t* minutes.
We want to know when the temperature is **100** degrees Fahrenheit, so we can set T(t)T(t) equal to **100**:
100=65+145e0.05t100 = 65 + 145e^{-0.05t}

STEP 4

Now, let's **isolate** that exponential term!
We can start by subtracting **65** from both sides of the equation:
10065=65+145e0.05t65100 - 65 = 65 + 145e^{-0.05t} - 6535=145e0.05t35 = 145e^{-0.05t}Then, we'll divide both sides by **145**:
35145=145e0.05t145\frac{35}{145} = \frac{145e^{-0.05t}}{145}35145=e0.05t\frac{35}{145} = e^{-0.05t}729=e0.05t\frac{7}{29} = e^{-0.05t}

STEP 5

To get that *t* out of the exponent, we can take the natural logarithm (ln) of both sides:
ln(729)=ln(e0.05t)\ln\left(\frac{7}{29}\right) = \ln\left(e^{-0.05t}\right)ln(729)=0.05t\ln\left(\frac{7}{29}\right) = -0.05t

STEP 6

Almost there!
Now we just need to divide both sides by **-0.05** to solve for *t*:
ln(729)0.05=0.05t0.05\frac{\ln\left(\frac{7}{29}\right)}{-0.05} = \frac{-0.05t}{-0.05}t=ln(729)0.05t = \frac{\ln\left(\frac{7}{29}\right)}{-0.05}t1.4270.05t \approx \frac{-1.427}{-0.05}t28.54t \approx 28.54So, it takes approximately **28.54 minutes** for the soup to cool to 100 degrees Fahrenheit.

STEP 7

We're given the formula β=10log(I1012)\beta = 10\log\left(\frac{I}{10^{-12}}\right), where β\beta is the sound intensity in decibels (dB) and II is the intensity in watts per square meter (W/m²).
We're told that β\beta is **80** dB, so let's plug that in:
80=10log(I1012)80 = 10\log\left(\frac{I}{10^{-12}}\right)

STEP 8

Let's **isolate** the logarithm term by dividing both sides by **10**:
8010=10log(I1012)10\frac{80}{10} = \frac{10\log\left(\frac{I}{10^{-12}}\right)}{10}8=log(I1012)8 = \log\left(\frac{I}{10^{-12}}\right)

STEP 9

Now, we can rewrite this logarithmic equation in exponential form.
Remember, if logb(a)=c\log_b(a) = c, then bc=ab^c = a.
In our case, the base is **10**:
108=I101210^8 = \frac{I}{10^{-12}}

STEP 10

Finally, multiply both sides by 101210^{-12} to solve for *I*:
1081012=I1012101210^8 \cdot 10^{-12} = \frac{I}{10^{-12}} \cdot 10^{-12}10812=I10^{8-12} = I104=I10^{-4} = ISo, the sound intensity is 10410^{-4} W/m².

STEP 11

The soup takes approximately **28.54 minutes** to cool to 100°F.
The intensity of the heavy traffic sound is **10410^{-4} W/m²**.

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