[5.2] Logarithms and Exponential Models

$$y= \log x \text{ if, and only if }10^y=x,$$ $$y= \ln x \text{ if, and only if }e^y=x.$$

Inverse functions:

$$10^{\log x}=x; \ \ \ \ \log(10^x)=x.$$ $$e^{\ln x}=x; \ \ \ \ \ln(e^x)=x.$$

Properties of logarithms (either common or natural)

$$\log(ab)=\log a+ \log b,$$ $$\log(a/b)=\log a- \log b,$$ $$\log(b^t)=t \log b.$$

These are related to these properties of exponents (either 10 or $e$)

$$10^a10^b=10^{a+b},$$ $$10^a/10^b=10^{a-b},$$ $$(10^b)^t=10^{bt}.$$

Examples

  • Solve $7e^{-3x}=49$ for $x$.
  • Simplify $e^{-2\ln(t/3)}$.
  • WileyPlus: 5.1 Question 30...

Exponential growth/decay

All of these equations can be used interchangeably to describe exponential growth (or decay):

  • $y(t)=a b^t$
  • $y(t)=a (1+r)^t$
  • $y(t)=ae^{rt}$ (slightly different $r$...)
  • $y(t)=a2^{t/\tau}$: where $\tau$ is the doubling time or (if negative) the half life.

Example 1

Let $f(t)=5(1.2)^t$. Find the annual growth rate, growth factor, continuous growth rate, and doubling time.

Growth rate

This is the $r$ in $y=a(1+r)^t$.

The $y$-intercept / initial value in the equation is $a=5$.

Apparently $(1+r)=(1.2)$, so $r=0.2$, which is 20% growth per year (if $t$ is in years).

Growth factor

This is the $b$ in $y=a b^t$. Oh, we can already see it: $b=1.2$.

Continuous growth rate

This is the $r$ in $y=a e^{rt}$.

Re-writing this as $$y=5e^{rt}=5(e^r)^t = 5 (1.2)^t,$$ Apparently $e^r=1.2$. So to find $r$ we solve this: $$\begineq e^r&=&1.2\\ \ln(e^r)&=&\ln(1.2)\\ r&=&0.182 \endeq $$ The continuous growth rate is 0.182 or 18.2 %

Doubling time

We know the initial function value $f(0)=a=5.

To find the doubling time, we need to find the $t$ that solves: $y(t)=2y(0)=2a=10=5(1.2)^t$ like this...

$$\begineq 10&=&5(1.2)^t\\ 2&=&(1.2)^t\\ \log 2&=&\log[(1.2)^t]=t\log 1.2\\ \frac{\log 2}{\log 1.2} &=& t\\ \frac{0.30103}{0.07918}=3.802 &=& t\\ \endeq $$ The doubling time is 3.802 (years, if $t$ is in years)

If we had used the natural log instead of the common log, we would have found: $$t=\frac{\ln 2}{\ln 1.2}=\frac{0.69314}{0.18232}=3.802$$

Example 2

Let $g(t)=5e^{-0.04 t}$. Find the annual growth rate, growth factor, continuous growth rate, and half life

Example 3 - population

The population of Earth is currently 7 billion and the expected doubling time is 26 years. Find a formula, the annual growth rate, growth factor, and continuous growth rate.

Example 4 - medication

A patient received an injection of 60 mg of a drug. After 3 hours there was still 40 mg in her blood streem. Find a formula, the annual growth rate, growth factor, and continuous growth rate.

Carbon dating

In 1991, the body of a man was found in melting snow in the Alps of Northern Italy. En examinatoin of the tissue sample revealed that 46% of the carbon-14 present in his body at the time of his death had decayed. The half-life of carbon-14 is approximately 5728 years. How long ago did this man die?