Mathematical Architects · Algebra II · Foundations
Get Ready: Module 5 — Exploring Exponentials & Logarithms
Everyone starts somewhere. Before we raise the building, let's pour the floor — one prerequisite skill at a time, at your own pace.
Foundations · Build the floor first
Module 5 asks you to think about exponentials and their mirror, logarithms.
That feels much easier once a few earlier skills are solid. This page walks through each one in plain
language, with a fully worked example and a couple of problems you can try.
How to use this page: Read each card. If a skill already feels easy, skim it and move on.
If it feels shaky, slow down, work the example with a pencil, then try the practice. There's no rush and
no test here — this is just a warm-up so the real module makes sense. You've got this.
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Skills to Build First
Six short warm-ups. Each one is a brick in the floor under Module 5.
1 Exponent Rules
An exponent is a shortcut for repeated multiplication: \(2^3\) means \(2 \times 2 \times 2 = 8\).
A handful of rules let you combine and simplify these powers. You need them in Module 5 because logarithms
are really just exponents wearing a different outfit — every log rule comes straight from an exponent rule.
Worked example
Simplify \(\dfrac{x^5 \cdot x^2}{x^3}\).
Multiply the top: when you multiply same-base powers, add exponents. \(x^5 \cdot x^2 = x^{5+2} = x^7\).
Divide: when you divide same-base powers, subtract exponents. \(\dfrac{x^7}{x^3} = x^{7-3} = x^4\).
Answer: \(x^4\).
Try it
a) Simplify \(3^2 \cdot 3^4\).
Show answer
Same base, so add exponents: \(3^{2+4} = 3^6 = 729\). Answer: \(3^6 = 729\).
b) Simplify \(\left(y^3\right)^2\) and write \(5^{-2}\) without a negative exponent.
Show answer
A power of a power multiplies exponents: \(\left(y^3\right)^2 = y^{3\cdot 2} = y^6\).
A negative exponent means "reciprocal": \(5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}\). Answers: \(y^6\) and \(\tfrac{1}{25}\).
An exponential function looks like \(y = a\cdot b^{x}\): the variable is up in the exponent.
When \(b > 1\) the output grows fast (growth); when \(0 < b < 1\) it shrinks (decay). Module 5 is built around
these curves, so being comfortable plugging in values and seeing the pattern is a huge head start.
Worked example
For \(y = 3 \cdot 2^{x}\), find \(y\) when \(x = 0,\ 1,\ 3\).
\(x=0\): anything to the 0 power is 1, so \(y = 3 \cdot 2^{0} = 3 \cdot 1 = 3\).
An inverse undoes something. Adding 5 is undone by subtracting 5; multiplying by 3 is undone
by dividing by 3. An inverse function swaps inputs and outputs — it sends you back where you came from.
This is the big idea of Module 5: a logarithm is the inverse of an exponential. It undoes "raise to a power."
Worked example
A machine does: "multiply by 2, then add 1." Undo it for an output of 11.
Reverse the order and reverse each step. The last thing done was "add 1," so first subtract 1: \(11 - 1 = 10\).
The first thing done was "multiply by 2," so now divide by 2: \(10 \div 2 = 5\).
Answer: the input was \(5\). Check forward: \(5 \times 2 = 10\), then \(10 + 1 = 11\). ✓
Try it
a) What undoes "raise to a power"? (This is the Module 5 punchline.)
Show answer
A logarithm. Just as subtraction undoes addition, a logarithm undoes an exponential. If \(2^x = 8\), then \(\log_2 8 = x\).
b) A machine does "subtract 4, then divide by 3." Undo it for an output of 2.
Show answer
Reverse, last step first: undo "divide by 3" by multiplying by 3: \(2 \times 3 = 6\). Then undo "subtract 4" by adding 4: \(6 + 4 = 10\). Answer: 10. Check: \(10 - 4 = 6\), \(6 \div 3 = 2\). ✓
Solving an equation means getting the variable by itself by doing the same thing to both sides
until it's alone. You'll lean on this constantly in Module 5 to solve exponential and logarithmic equations —
the variable just happens to be hiding in an exponent.
Worked example
Solve \(3x + 4 = 19\).
Undo the +4 by subtracting 4 from both sides: \(3x = 19 - 4 = 15\).
Undo the \(\times 3\) by dividing both sides by 3: \(x = \dfrac{15}{3} = 5\).
\(f(x)\) is just a name for a rule, read "f of x." It does not mean \(f\) times \(x\) — it means
"put \(x\) into the rule \(f\)." Module 5 writes exponentials and logarithms this way, like \(f(x) = 2^{x}\),
so reading and evaluating this notation smoothly keeps you from getting tripped up.
Worked example
Given \(f(x) = 2x + 5\), find \(f(3)\).
\(f(3)\) means "replace every \(x\) with 3." So \(f(3) = 2(3) + 5\).
The coordinate plane is the grid where we draw graphs. A point is an ordered pair \((x, y)\): go across
for \(x\), then up or down for \(y\). Module 5 graphs exponentials and logs and shows they're mirror
images across the line \(y = x\) — so plotting points and reading them confidently is key.
Worked example
Plot \((3, 8)\), then find its mirror across the line \(y = x\).
Plot \((3, 8)\): start at the origin, move 3 right, then 8 up.
Mirror across \(y = x\): reflecting over that line simply swaps the coordinates, giving \((8, 3)\).
Why it matters: \((3,8)\) sits on \(y = 2^{x}\) and its swap \((8,3)\) sits on \(y = \log_2 x\). The swap is the inverse.
Try it
a) What is the mirror of the point \((2, 5)\) across the line \(y = x\)?
Show answer
Swap the coordinates: \((5, 2)\). Answer: \((5, 2)\).
b) In which direction do you move first for the point \((-4, 1)\)?
Show answer
The \(x\)-value is \(-4\), so move 4 to the left first, then 1 up. Answer: left 4, then up 1.