M151Q Precalculus · Unit 2: Powers, exponentials, logarithms · on Exam 2
MFG 5.2–5.3 Logarithmic functions and the natural base
Study notes for Logarithmic functions and the natural base in M151Q Precalculus at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in StudyCat.
Big ideas
- $y = \log_b x$ is the inverse of $y = b^x$: domain $(0,\infty)$, range all reals, vertical asymptote $x = 0$, passes through $(1,0)$ and $(b,1)$. Increasing for $b>1$, slowly.
- $e \approx 2.718$ is the natural base; $\ln x = \log_e x$. $e^x$ and $\ln x$ are inverses: $e^{\ln x} = x$, $\ln e^x = x$.
- Continuous growth $A = A_0e^{kt}$ with continuous rate $k$; the ordinary growth factor is $b = e^k$, so $k = \ln b$. Doubling time $= \ln 2/k$.
- Solve exponential equations with $\ln$: $e^{0.3t} = 5 \Rightarrow t = \ln 5/0.3$. Solve log equations by rewriting in exponential form; reject solutions that make any log argument $\le 0$.
Key formulas
Natural exponential and log
$$y = e^x \iff x = \ln y,\qquad \ln e = 1,\quad \ln 1 = 0$$
Continuous growth
$$A = A_0e^{kt},\qquad b = e^k,\qquad t_{\text{double}} = \frac{\ln 2}{k}$$
Log equations
$$\log_b(\text{expr}) = c \Rightarrow \text{expr} = b^c$$
Worked example
Solve $\ln(2x - 3) = 2$ and find the doubling time for $A = 300e^{0.045t}$.
Show the solution
$2x - 3 = e^2 \Rightarrow x = \dfrac{e^2 + 3}{2} \approx 5.19$ (argument $2x-3 = e^2 > 0$ ✓). Doubling time $= \ln 2/0.045 \approx 15.4$ years.
Common mistakes
- Keeping a solution that makes a log argument negative or zero.
- Mixing up ln (base e) and log (base 10) on a calculator.
Exam tipWhen a problem says "continuous", the model is e^{kt}. When it says "per year", the model is b^t. Convert with k = ln b.
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