FunctionsMAT 102

Example4.B.1Properties of Logarithms

Suppose you are given these two facts:

• \(\log_4 5 = 1.16\)
• \(\log_4 10 = 1.66\)

Then we can use the laws of logarithms to conclude that:

• \(\log_4 (50) = \log_4 5 + \log_4 10 = 1.16 + 1.66 = 2.82\)
• \(\log_4 (2) = \log_4 10 - \log_4 5 = 1.66 - 1.16 = 0.5\)
• \(\log_4 (100,000) = 5\log_4 10 = 5(1.66) = 8.3\checkmark\)

Example4.B.2Properties of Logarithms

Expand the following using the rules of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers.

1. \(\log_2\left( \dfrac{8}{x}\right)\)

   Solution: To expand \(\log_2\left( \frac{8}{x}\right)\text{,}\) we use the Quotient Rule:

   \begin{align*} \log_2\left( \dfrac{8}{x}\right)\amp = \log_2(8) - \log_2(x)\amp\amp\text{Quotient Rule}\\ \amp = 3 - \log_2(x)\amp\amp\text{Since \(2^3 = 8\) we can simplify the }\log_2(8)=3 \end{align*}

2. \(\ln (x^2y)\)

   Solution: To expand this expression, we need both the Product Rule and the Power Rule:

   \begin{align*} \ln (x^2y)\amp = \ln(x^2) + \ln(y)\amp\amp\text{Product Rule}\\ \amp = 2\ln(x) + \ln(y)\amp\amp\text{Power Rule} \end{align*}

3. \(\log \left( \dfrac{\sqrt{x}}{x+3} \right)\)

   Solution: The Quotient Rule and Power Rule will do the trick for this one:

   \begin{align*} \log \left( \dfrac{\sqrt{x}}{x+3} \right)\amp = \log(\sqrt{x}) - \log(x+3)\amp\amp\text{Quotient Rule}\\ \amp = \log(x^{1/2}) - \log(x+3)\amp\amp\text{Rewrite the square root as an exponent}\\ \amp = \frac{1}{2}\log(x) - \log(x+3)\amp\amp\text{Power Rule} \end{align*}

   Note that we cannot change \(\log(x+3)\) by any log rule since it is not a product nor a quotient.

4. \(\log(100A^{0.3}B)\)

   Solution:

   \begin{align*} \log(100A^{0.3}B)\amp = \log(100) + \log(A^{0.3})+\log(B)\amp\amp\text{Product Rule}\\ \amp = 2 + \log(A^{0.3})+\log(B)\amp\amp\text{Simplify }\log(100) = 2\text{ since }10^2=100\\ \amp = 2 + 0.3\log(A)+\log(B)\text{Power Rule} \end{align*}

5. \(\ln(x^2 - 5y)\)

   Solution: This one does not simplify since it is not a product nor quotient.\(\checkmark\)

Example4.B.3Log Rules

Use the rules of logarithms to write the following as a single logarithm.

1. \(\log(x) + \log(y) - \log(5)\)

   Solution:

   \begin{align*} \log(x) + \log(y) - \log(5)\amp = \log(xy) - \log(5)\amp\amp\text{Product Rule to combine }\log(x) + \log(y)\\ \amp = \log\left( \dfrac{xy}{5} \right)\amp\amp\text{Quotient Rule} \end{align*}

2. \(2\ln(x-1) - \ln(x+1)\)

   Solution:

   \begin{align*} 2\ln(x-1) -\ln(x+1)\amp = \ln(x-1)^2 -\ln(x+1)\amp\amp\text{Power Rule}\\ \amp = \ln\left(\dfrac{(x-1)^2}{x+1}\right)\amp\amp\text{Quotient Rule} \end{align*}

3. \(\ln(x)\ln(x+1)\)

   Solution: This expression cannot be combined into one log since it is not a sum or difference of two logarithms.

4. \(\dfrac{\log(5t)}{\log(t-3)}\)

   Solution: Like the previous example, this expression cannot be combined into one logarithm since it is not the sum or difference of two logarithms.\(\checkmark\)

Example4.B.4Earthquakes

In the western United States the area \(A\) (in \(\mbox{mi}^2\)) affected by an earthquake is related to the magnitude \(R\) of the quake by the formula:

1. If \(5 \mbox{ mi}^2\) of land are affected by an earthquake, what was the magnitude of the quake?

   Solution: Use \(A = 5\) in the formula to find \(R\text{:}\)

   \begin{align*} R \amp = 2.3 \log(5 + 3000) - 5.1\\ \amp = 2.3 \log(3005) - 5.1\\ \amp \approx 2.3 (3.47784) - 5.1\\ \amp \approx 2.89903 \end{align*}

   \(R\) is about a magnitude \(3\) earthquake.

2. How many square miles will be affected if a magnitude \(5\) earthquake hits?

   Solution: This time we know \(R\) but not \(A\text{.}\)

   \begin{align*} 5\amp = 2.3 \log(A + 3000) - 5.1\amp\amp\\ 5 + 5.1 \amp = 2.3 \log(A + 3000)\amp\amp\\ \frac{10.1}{2.3}\amp = \log(A + 3000)\amp\amp\text{Switch to exponential form}\\ A + 3000 \amp = 10^{10.1/2.3}\amp\amp\\ A\amp=10^{10.1/2.3} - 3000\amp\amp\\ \amp\approx 21,620.924 \text{ mi}^2\checkmark \end{align*}