The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 22 (Intermediate Algebra) Logarithms, Compound Interest, and Trig Functions

Logarithms, Compound Interest, and Trig Functions – Lecture 22

Overview

This lecture closes out the course by sweeping through the final three topics of the book: logarithms, interest formulas, and right-triangle trigonometry. A logarithm is introduced as simply another way of writing an exponent, giving us a tool for working with exponents directly. We then apply exponential growth to money through the compound-interest and continuous-compounding formulas. Finally, we build the three basic trig ratios (sine, cosine, tangent) from right triangles, use them to find missing sides, and use their inverses to recover missing angles. The unifying idea is the relationship between a quantity and its exponent or ratio, and how to move back and forth between the two.

Key Concepts

Logarithms as exponents

• A logarithm answers the question: to what power must I raise the base to get this number? A logarithm takes a number and spits out an exponent; an exponential takes an exponent and spits out a number. They are inverse operations.
• Definition: $\log_a b = c$ is logically equivalent to $a^c = b$. Here $a$ is the base, $c$ is the exponent (always the output), and $b$ is the number.
• To rewrite between forms, identify the base, exponent, and number, then place them in the correct slots. Example: $\left(\tfrac{2}{3}\right)^4 = \tfrac{16}{81}$ becomes $\log_{2/3}\tfrac{16}{81} = 4$.
• Evaluating a log: set it equal to $x$, rewrite as an exponential, then express both sides with a common base. E.g. $\log_2 64 = x \Rightarrow 2^x = 2^6 \Rightarrow x = 6$.
• Common base: $\log_{125} 5 = x \Rightarrow 125^x = 5 \Rightarrow 5^{3x} = 5^1 \Rightarrow x = \tfrac{1}{3}$.
• Negative results: $\log_3 \tfrac{1}{27} = x \Rightarrow 3^x = 3^{-3} \Rightarrow x = -3$.
• Common log (base 10): when no base is written, the base is assumed to be $10$. So $\log x$ means $\log_{10} x$. E.g. $\log x = -2 \Rightarrow 10^{-2} = x \Rightarrow x = \tfrac{1}{100}$.

Simple vs. compound interest

• Simple interest charges interest only on the original principal each year, so the balance grows by the same fixed amount every period.
• Compound interest charges interest on the interest already accrued, so the balance grows faster and faster. Over long horizons the gap between simple and compound is enormous.
• Compound interest formula: $A = P\left(1 + \frac{r}{n}\right)^{nt}$ where $P$ = starting principal, $r$ = annual interest rate (as a decimal), $n$ = number of compounding periods per year, and $t$ = number of years. Common values of $n$: quarterly $= 4$, monthly $= 12$, weekly $= 52$, daily $= 365$.
• As $n$ grows (yearly → monthly → daily → every second), the balance increases but the gains shrink, approaching a limit called continuous compounding.
• Continuous compounding formula: $A = Pe^{rt}$ where $e \approx 2.71828\ldots$ is a fixed constant on your calculator (like $\pi$). There is no $n$ because compounding happens continuously.

Right-triangle trigonometry (SOH-CAH-TOA)

• Orient the triangle with the angle $\theta$ in the bottom-left corner. The longest side, opposite the right angle, is the hypotenuse ($h$). The side touching the angle is adjacent ($a$); the side across from it is opposite ($o$).
• The three trig functions are ratios: $\sin\theta = \frac{o}{h}, \qquad \cos\theta = \frac{a}{h}, \qquad \tan\theta = \frac{o}{a}$
• Each function takes an angle and returns a ratio of side lengths. The ratio depends only on the angle, not on the triangle's size — scaling a triangle leaves the ratios unchanged.
• Finding a side: when you know the angle and one side, pick the function relating the side you want to the side you have, then solve algebraically.

Inverse trig functions

• The inverse trig functions undo the trig functions: you give them a ratio and they return the angle.
• $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ are used when you know two sides and want the angle. E.g. $\sin A = 0.5 \Rightarrow A = \sin^{-1}(0.5) = 30^\circ$.
• Solving a triangle means finding all three angles and all three sides. Use the angle sum ($90^\circ + \text{the two acute angles} = 180^\circ$), the trig ratios, and optionally the Pythagorean theorem $a^2 + b^2 = h^2$. There are several valid routes to the same answer.

Worked Example

(a) Compound interest. A $25,000 car loan at $6.5\%$ interest compounded quarterly, with no payment for $5$ years. Find the balance.

Identify the variables: $P = 25000$, $r = 0.065$, $n = 4$, $t = 5$.

Taking five years off from paying adds roughly $9,500 to what is owed.

(b) Solving a right triangle. A right triangle has one acute angle of $35^\circ$ and the side adjacent to it equal to $5$. Find the missing acute angle, the opposite side $x$, and the hypotenuse $h$.

As a check, the Pythagorean theorem gives $h = \sqrt{5^2 + 7.14^2} \approx 8.72$, matching.

Summary

• A logarithm is an exponent in disguise: $\log_a b = c \iff a^c = b$; evaluate by rewriting with a common base, and remember $\log$ with no base means base $10$.
• Compound interest charges interest on interest, so it outgrows simple interest dramatically over time.
• Use $A = P\left(1 + \frac{r}{n}\right)^{nt}$ for periodic compounding (match $n$ to quarterly/monthly/weekly/daily) and $A = Pe^{rt}$ for continuous compounding.
• In a right triangle, SOH-CAH-TOA gives $\sin\theta = \frac{o}{h}$, $\cos\theta = \frac{a}{h}$, $\tan\theta = \frac{o}{a}$; orient the angle to the bottom-left and pick the ratio that links the sides you have and want.
• Inverse trig functions ($\sin^{-1}, \cos^{-1}, \tan^{-1}$) turn a ratio of two known sides back into the angle.
• To "solve a triangle," combine the $180^\circ$ angle sum, the trig ratios, and the Pythagorean theorem; multiple correct paths reach the same sides and angles.