Logarithms [1.6]

But first, a reminder to both you and me... Preparation: From our syllabus...

You are expected to come prepared. This means that you should have read the sections to be covered and should be prepared to ask questions about the reading and the problems you are asked to do.

Research has shown that "active learning" in which students spend class time engaging with material, is more effective than hearing about content. Instead of imparting information in lectures, class activities will be focussed on applications and implications. And this means you need to get enough of the content ahead of time to engage with it.

Characteristics of all logarithms

  1. $\log(AB)=\log A + \log B$
  2. $\log\left(\frac AB\right)=\log A -\log B$
  3. $\log (A^p)=p\log A$
  4. Also $\log 1=0$
  5. $\color{red}{\log_{10} 10=1}$ and $\color{blue}{\ln{e}=1}$
    where $e$ is Euler's constant, and has a value of $e\approx 2.72$.
  6. $\log x$ is not defined if $x$ is negative or 0.

The logarithm, base 10 of $x$, written $\log_{10} x=c$, is the power of $10$ needed to get $x$: $$10^c=x \ \ \Leftrightarrow \ \ \ \log_{10} x = c$$

This also means... $$\log_{10}10^c = c \ \ \text{ and } \ \ 10^{\log_{10}x}=x $$

The logarithm, base e of $x$, written $\log_e x\equiv\ln x=c$, is the power of $e$ needed to get $x$: $$e^c=x \ \ \Leftrightarrow \ \ \ \ln x = c$$

This also means... $$\ln e^c = c\ \ \text{ and }\ \ e^{\ln x}=x$$

Problems

  1. Solve this equation for $t$: $$200=30e^{0.15t}$$

    Start by dividing both sides of the equation above by 30: $$\begineq \frac{200}{30}=6.666...&= e^{0.15t}\\ \ln(6.666)&=\ln(e^{0.15t}) = 0.15t\\ \frac{\ln(6.666)}{0.15}=\color{blue}{12.65...}&=\color{blue}{t}. \endeq $$

  2. Solve this equation for $x$: $$y=e+2^x.$$

    Subtracting $e$ from both sides of this equation: $$\begineq y-e&=2^x\\ \ln(y-e)&=\ln(2^x) = x\ln(2)\\ \color{blue}{\frac{\ln(y-e)}{\ln(2)}}&= \color{blue}{x}\\ \endeq $$

    Watch out for this...There's no easy way to simplify $\ln(y-e)$:

    It is so tempting to write $\ln(5-2)=\ln 5 - \ln 2$. But

    • According to #2 of "Characteristics of logarithms" (above), $\ln 5 - \ln 2=\ln(5/2)$,
    • and $\ln(5-2)=\ln(3)$
    • and clearly $5/2 \neq 3$.
  3. Convert the function $P=100(1.07)^t$ into an equivalent equation of the form $P=P_0e^{kt}$


    $P_0\equiv P(t=0)=100(1.07)^0=100.$

    So that leaves $(1.07)^t = e^{kt}$.
    But since $e^{kt}=(e^k)^t$, we have $(1.07)^t=(e^k)^t$ which means $$1.07=e^k.$$ From the definition of the natural log, $e^k=c$ means $k=\ln c$, so $$k=\ln 1.07 = 0.06766$$ Putting this all together: $$\color{blue}{P=100e^{0.06766t}}.$$

  4. Convert the function $P=750e^{0.04t}$ into the form $P=P_0a^t$.


    Since $(e^k)^t=e^{kt}$, we can write $e^{0.04t}$ as $(e^{0.04})^t=(1.0408)^t$, so, $$\color{blue}{P=750(1.0408)^t}$$