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This lesson builds on the log properties and equation-solving from the algebra curriculum, applying them to growth models and magnitude comparisons that show up specifically on the AMC 12.

Growth and decay models

A(t) = A_0 \, e^{rt}

where $A_0$ is the initial amount, $r$ is the (continuous) growth rate (negative for decay), and $t$ is time. A useful derived fact: the doubling time for growth (or half-life for decay) comes directly from setting $A(t)/A_0 = 2$ (or $\tfrac12$ ) and solving:

t_{\text{double}} = \dfrac{\ln 2}{r}

Comparing exponential magnitudes

To compare two large exponential expressions without a calculator, take the log of both sides — this turns a comparison of huge numbers into a comparison of ordinary-sized products, which is often easy to estimate directly.

Worked example

Problem: Which is larger, $2^{300}$ or $3^{200}$ ?

Solution: Take $\log$ (base 10, or any consistent base) of both:

\log(2^{300}) = 300\log2 \approx 300(0.301) = 90.3

\log(3^{200}) = 200\log3 \approx 200(0.477) = 95.4

Since $95.4 > 90.3$ , we conclude $3^{200} > 2^{300}$ .

Practice problems

1.A population grows according to $P(t)=P_0e^{0.05t}$ . Find the doubling time in terms of $\ln2$ .

2.Which is larger: $5^{100}$ or $3^{150}$ ? (Use $\log5\approx0.699, \log3\approx0.477$ .)

3.A radioactive substance has a half-life of 10 years. Express the decay rate $r$ in terms of $\ln2$ .

4.Estimate $\log_{10}(2^{50})$ using $\log_{10}2\approx0.301$ , and use it to find the number of digits in $2^{50}$ .

Logarithms & Exponential Growth

Growth and decay models

Comparing exponential magnitudes

Practice problems