Precalculus: Lecture Notes

Section 3.2 Rules for Logarithms

Let's not go and ruin it by thinking too much. — Clinton Eastwood, Jr., American actor, film director, producer, and composer, 1930–

Problem: A day after your graduation you realize that the total amount of your student loan is \(A=\$20,000\) at \(i=3\%\) annual interest compounded monthly. If your payment is \(P=\$300\) a month, how long will it take to repay the loan?

Reminder: For any \(a\gt 0\text{,}\) \(a\not= 1\text{,}\) any \(x\gt 0\text{,}\) and any \(y\in \mathbb{R}\)

The Properties of Logarithms. For any positive real numbers \(x,y\) and any positive real number \(a\text{,}\) \(a\not= 1\text{:}\)

• \begin{equation*} \log_a(xy)=\log _a(x)+\log _a(y) \end{equation*}

  Proof: Let \(\log_a(xy)=u\text{,}\) \(\log _a(x)=v\text{,}\) and \(\log _a(y)=w\text{.}\) Then, by definition, \(xy=a^u\text{,}\) \(x=a^v\text{,}\) and \(y=a^w\text{.}\) It follows that

  \begin{equation*} a^u=xy=(a^v)(a^w)=a^{v+w}\Leftrightarrow u=v+w\Leftrightarrow \log_a(xy)=\log _a(x)+\log _a(y). \end{equation*}

  Example 3.2.1. Logarithm of a product.

  Solve

  \begin{equation*} \log _3(x)=\log _3(2)+\log _3(x^2-3) \end{equation*}

  Solution

  Observe that the domain of the given equation is the set \(\{x: x\gt 0\text{ and } x^2\gt 3\}=\{x:x\gt \sqrt{3}\}=(\sqrt{3},\infty)\text{.}\)

  Hence, for \(x\in (\sqrt{3},\infty)\text{,}\)

  \begin{equation*} \log _3x=\log _3(2)+\log _3(x^2-3)\Leftrightarrow \log _3x=\log _3(2(x^2-3))\Leftrightarrow x=2x^2-6\Leftrightarrow 2x^2-x-6=0. \end{equation*}

  This is a quadratic equation and its solutions are given by

  \begin{equation*} x=\frac{1\pm\sqrt{1+48}}{4}=\frac{1\pm 7}{4}. \end{equation*}

  We observe that only \(x=2\) belongs to the domain of the given equation. Hence, \(x=2\) is the only solution.

• \begin{equation*} \log_a\left(\frac{x}{y}\right)=\log _a(x)-\log _a(y) \end{equation*}

  Proof: We observe that

  \begin{equation*} \log_a\left(\frac{x}{y}\right)+\log _a(y)=\log_a\left( \frac{x}{y}\cdot y\right)=\log_a(x). \end{equation*}

  Hence, \(\displaystyle \log_a\left(\frac{x}{y}\right)=\log _a(x)-\log _a(y)\text{.}\)

  Example 3.2.2. Logarithm of a quotient.

  Solve

  \begin{equation*} \log (x-2)-\log (x)=3 \end{equation*}

  Solution

  We note that the domain of the given equation is the set \(\{x:x-2\gt 0\text{ and }x\gt 0\}=\{x:x\gt 2\}=(2,\infty)\text{.}\)

  It follows that for \(x\in (2,\infty)\text{,}\)

  \begin{equation*} \log (x-2)-\log x=3\Leftrightarrow \log\left(\frac{x-2}{x}\right)=3\Leftrightarrow \frac{x-2}{x}=10^3\Leftrightarrow x-2=1000x\Leftrightarrow 999x=-2 \end{equation*}

  But \(\displaystyle x=-\frac{2}{999}\lt 0\) and we conclude that the given equation has no solution.

• \begin{equation*} \log _a {\left(x^r\right)}=r\log _a(x) \end{equation*}

  Proof: Let \(t=\log _a {\left(x^r\right)}\text{.}\) Then, for \(r\not= 0\text{,}\)

  \begin{equation*} t=\log _a {\left(x^r\right)} \Leftrightarrow x^r=a^t\Leftrightarrow x=a^{t/r}\Leftrightarrow \log_a(x)=t/r\Leftrightarrow t=r\log_a(x). \end{equation*}

  Hence,

  \begin{equation*} \log _a {\left(x^r\right)}=r\log _a(x). \end{equation*}

  Example 3.2.3. Logarithm of a power.

  Solve

  \begin{equation*} 2^{x-2}=3^x \end{equation*}

  Solution

  We note

  \begin{equation*} 2^{x-2}=3^x\Leftrightarrow \log \left(2^{x-2}\right)=\log \left(3^x\right)\Leftrightarrow (x-2)\log (2)=x\log (3)\Leftrightarrow x\log (2)-2\log (2)=x\log (3) \Leftrightarrow \end{equation*}

  \begin{equation*} x\log (2)-x\log (3)=2\log (2)\Leftrightarrow (\log (2)-\log (3))x=2\log (2)\Leftrightarrow x=\frac{2\log (2)}{\log (2)-\log (3)}. \end{equation*}
• \begin{equation*} \log _a(1)=0 \end{equation*}

  Proof: Note that

  \begin{equation*} \log _a(1)=0\Leftrightarrow a^0=1. \end{equation*}

  Hence,

  \begin{equation*} \log _a(1)=0. \end{equation*}

  Example 3.2.4. Logarithm of \(1\).

  Solve

  \begin{equation*} \log_2(4x^2-12x+10)=0 \end{equation*}

  Solution

  From

  \begin{equation*} \log_2(4x^2-12x+10)=0\Leftrightarrow \log_2(4x^2-12x+10)=\log (1)\Rightarrow 4x^2-12x+10=1\Leftrightarrow \end{equation*}

  \begin{equation*} x^2-12x+9=0\Leftrightarrow (2x-3)^2=0\Leftrightarrow x=\frac{2}{3}. \end{equation*}

  We check if \(\displaystyle \frac{2}{3}\) is in the domain of the given equation:

  \begin{equation*} 4\cdot \left(\frac{3}{2}\right)^2-12\cdot \frac{3}{2}+10=1\gt 0. \end{equation*}

  Hence \(x=\displaystyle \frac{2}{3}\) is the solution of the given equation.

• \begin{equation*} \log _a{\left(a^x\right)}=x \end{equation*}

  Proof: Observe

  \begin{equation*} \log _a{\left(a^x\right)}=x\Leftrightarrow a^x=a^x. \end{equation*}

  Example 3.2.5. Logarithm of an exponent.

  Solve

  \begin{equation*} \log \left(10^{x^2}\right)+\log \left(10^{-12}\right)=\log \left(10^x\right) \end{equation*}

  Solution

  Note that the domain of the given equation is the set of all real numbers.

  From

  \begin{equation*} \log \left(10^{x^2}\right)=x^2, \ \log \left(10^{-12}\right)=-12, \text{ and }\log \left(10^x\right)=x \end{equation*}

  we obtain that the given equation is equivalent to the equation

  \begin{equation*} x^2-12=x\Leftrightarrow x^2-x-12=0\Leftrightarrow (x-4)(x+3)=0. \end{equation*}

  Hence the solutions of the given equation are \(x=-3\) and \(x=4\text{.}\)

• \begin{equation*} \log _a (N)=\frac{\log _b(N)}{\log _b (a)} \end{equation*}

  Proof: Let \(t=\log_a(N)\text{.}\) Then

  \begin{equation*} t=\log_a(N)\Leftrightarrow N=a^t\Leftrightarrow \log_b(N)=\log_b(a^t)\Leftrightarrow \log_b(N)=t\log_b(a) \Leftrightarrow t=\frac{\log _b(N)}{\log _b a}. \end{equation*}
  \begin{equation*} \log _a (N)=\frac{\log _b(N)}{\log _b (a)}. \end{equation*}

  Example 3.2.6. Change of base formula.

  Solve

  \begin{equation*} \log_{x^2}(10)+\log_{x}(100)=\log_2 (1000) \end{equation*}

  Solution

  Note that the domain of the given equation is the set \(\{x:x\gt 0,x\not= 1\}\text{.}\)

  From

  \begin{equation*} log_{x^2}(10)=\frac{\log (10)}{\log \left(x^2\right)}=\frac{1}{2\log (x)}, \ \log_{x}(100)=\frac{\log (100)}{\log (x)}=\frac{2}{\log (x)}, \text{ and }\log_2 (1000)=\frac{\log (1000)}{\log (2)}=\frac{3}{\log (2)} \end{equation*}

  we obtain that the given equation is equivalent to the equation

  \begin{equation*} \frac{1}{2\log (x)}+\frac{2}{\log (x)}=\frac{3}{\log (2)}. \end{equation*}

  It follows

  \begin{equation*} \frac{1}{2\log (x)}=\frac{3}{5\log (2)}\Leftrightarrow \log x=\frac{5\log (2)}{6}\Leftrightarrow x=10^{\frac{5\log (2)}{6}}. \end{equation*}

  Hence the solutions of the given equation is \(\displaystyle x=10^{\frac{5\log (2)}{6}}\text{.}\)

MUST KNOW!

Whenever are all of the listed numbers defined:

1. \(\displaystyle \log_a(xy)=\log _a(x)+\log _a(y)\)

2. \(\displaystyle \displaystyle \log_a\left(\frac{x}{y}\right)=\log _a(x)-\log _a(y)\)

3. \(\displaystyle \log _a {\left(x^r\right)}=r\log _a(x)\)

4. \(\displaystyle \log _a(1)=0\)

5. \(\displaystyle log _a{\left(a^x\right)}=x\)

6. \(\displaystyle \displaystyle \log _a (N)=\frac{\log _b(N)}{\log _b (a)}\)

Exponents and logarithms … are part of life for many past and present students in Canada:

Suppose that the amount of \(A\) dollars has been borrowed at the interest per the payment period of \(i\) percent (written as a decimal fraction). Suppose that the borrower pays \(P\) dollars at the beginning of the each payment period.

Recall the following facts:

1. The loan balance \(B\) after \(n\) payments have been made is given by

   \begin{equation*} B=A(1+i)^n-\frac{P}{i}\left[(1+i)^n-1\right]. \end{equation*}

2. Payment \(P\) needed to repay the amount \(A\) in \(N\) payments at the interest \(i\) per the payment period is given by

   \begin{equation*} P=\frac{iA}{1-(1+i)^{-N}}. \end{equation*}

3. Number \(N\) of payments\(P\) needed to repay the amount \(A\) at the interest \(i\) per the payment period is given by

   \begin{equation*} N=\frac{-\log(1-iA/P)}{\log(1+i)} \end{equation*}