High School

1. Expand the logarithm as much as possible. Use the properties of logarithms to rewrite the expression as a sum, difference, or product of logs.

\[
\log(x^2 y^{-1})
\]

2. Condense the expression to a single logarithm using the properties of logarithms.

\[
\log(x) - \log(y) + 6\log(z)
\]

3. Use properties of logarithms to evaluate without using a calculator:

\[
\log_{10}(64) + \log_{10}(2) + 3\log_{10}(4)
\]

4. Use logarithms to solve:

\[
e^{2x} - e^x - 72 = 0
\]

5. Atmospheric pressure \( P \) in pounds per square inch is represented by the formula:

\[
P = 14.7e^{-0.21x}
\]

where \( x \) is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.544 pounds per square inch? (Hint: there are 5,280 feet in a mile.)

The mountain is ______ feet high.

6. A tumor is injected with 0.8 grams of Iodine-125, which has a decay rate of 1.15% per day. Write an exponential model representing the amount of Iodine-125 remaining in the tumor after \( t \) days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days.

\[
A(t) =
\]

There will be ______ grams of Iodine-125 after 60 days. (Round to nearest tenth)

7. A formula for calculating the magnitude of an earthquake is \( M = \) (uses the common (base 10) logarithm). This is called the Moment Magnitude Scale (MMS), an alternative to the more well-known Richter Scale. One earthquake has a magnitude of 3.9 on the MMS. If a second earthquake has 700 times as much energy as the first, find the magnitude of the second quake.

The magnitude of the second earthquake is ______. (Round to hundredth)

Answer :

1. To expand the logarithm using the properties of logarithms, we have:

log(x°y-1) = log(x^y/y)log(x°y-1) = log(x^y) - log(y)log(x°y-1) = ylog(x) - log(y)

Therefore, log(x°y-1) can be rewritten as ylog(x) - log(y).2. To evaluate the expression without using a calculator, we have:

log(64) / log(2) + 3log(4) = 6log(64) / log(2) + log(4^3) = 6log(2^6) / log(2) + 3log(2^2)

= 6(6) / 1 + 3(2)

= 36 + 6

= 42

Therefore, log(64) / log(2) + 3log(4) = 42.3.

To solve e^(2x) – e^x – 72 = 0, we can substitute y = e^x to obtain y^2 – y – 72 = 0(y – 9)(y + 8) = 0Therefore, y = 9 or y = -8Substituting back to obtain x:When y = 9, e^x = 9, so x = ln(9)When y = -8, e^x = -8, which is not possible Therefore, x = ln(9).4. To find the height of a mountain with an atmospheric pressure of 8.544 pounds per square inch, we can substitute P = 8.544 into the formula P = 14.7e^(-0.21x) to obtain:8.544 = 14.7e^(-0.21x)ln(8.544 / 14.7) = -0.21xln(8.544 / 14.7) / -0.21 = x Therefore, x is approximately 16,515 feet, so the mountain is approximately 16,515 feet high.5. To find the exponential model representing the amount of lodine-125 remaining in the tumor after t days, we can use the formula A(t) = A0(1 – r)^t, where A0 is the initial amount of lodine-125 and r is the decay rate expressed as a decimal. Since 1.15% = 0.0115, we have:A(t) = 0.8(1 – 0.0115)^tA(t) = 0.8(0.9885)^t To find the amount of lodine-125 remaining after 60 days, we substitute t = 60 into the formula to obtain:A(60) = 0.8(0.9885)^60A(60) ≈ 0.447 grams.

Therefore, the amount of lodine-125 remaining after 60 days is approximately 0.447 grams.6. To find the magnitude of the second earthquake, we use the fact that the energy of an earthquake is proportional to 10^(1.5M), where M is the magnitude on the Richter Scale. Since the second earthquake has 700 times as much energy as the first, we have:

10^(1.5M2) / 10^(1.5M1)

= 70010^(1.5M2 – 1.5M1)

= 700log(10^(1.5M2 – 1.5M1))

= log(700)1.5M2 – 1.5M1

= log(700)M2 – M1

= log(700) / 1.5M2

= M1 + log(700) / 1.5

Since the first earthquake has magnitude 3.9 on the MMS, we have:M2 = 3.9 + log(700) / 1.5M2 ≈ 5.46Therefore, the magnitude of the second earthquake is approximately 5.46 (rounded to the hundredth).

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