Exponents are just shorthand. Instead of writing $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$, we write $2^5$ and save ourselves a bunch of ink. There are only a handful of rules in this whole section, and every single one of them comes from that one idea of repeated multiplication. The catch is that the ISEE knows exactly which rules students mix up, and it builds the wrong answers around those mix-ups. So let's learn the rules and the traps. The practice is mixed into the lesson: each question sits right after the idea it needs, and the mixed set at the bottom takes that hint away.
Here's the whole idea in one picture. Watch the copies pile up, and watch what the little number up top is really counting.
First, some names for the parts. Every exponent problem has a number being multiplied and a number counting how many times.
In an expression $b^{x}$, $b$ is the base and $x$ is the exponent.
The whole thing, base and exponent together, has a name too.
For a counting number $x$, the value of $b^{x}$ is a power, meaning "multiply $x$ copies of $b$ together" (with $b^{1} = b$). The expression $b^{x}$ is read "$b$ to the power of $x$."
Let's use those names right away on the number from the picture.
What is the value of $2^{5}$?
The base is $2$ and the exponent is $5$, so multiply five $2$'s together:
$$2^{5} = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$$
Not $10$! That's $2 \times 5$, and it's the first trap answer every test writer reaches for.
Your turn. Count the copies, multiply them out, and don't let the base and the exponent multiply each other.
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Now for the rules. Each one is just a shortcut for counting how many copies of the base you end up with, so if you ever forget one, write out the copies and count.
Same base (any $b \neq 0$), three shortcuts:
$$b^{x}\cdot b^{y}=b^{x+y}, \qquad \frac{b^{x}}{b^{y}}=b^{x-y}, \qquad \left(b^{x}\right)^{y}=b^{x\cdot y}$$
Here's what's really going on. Every power is a row of $b$ tiles. A positive exponent puts tiles on top of the fraction bar and a negative exponent puts them underneath. Any $b$ on top cancels a $b$ on the bottom (because $\frac{b}{b} = 1$), and whatever survives is your answer. Try all three operations, then go for the missions.
See it? Multiplying piles the tiles together, so the exponents add. Dividing sends the second batch of tiles to the other side of the bar, so they subtract. And a power of a power just makes copies of the whole row, so the exponents multiply. The hard part on the test is telling which of the three you're looking at, so sort a few.
Once you can tell them apart, the ISEE turns them around and hides the exponent instead.
If $(n^{k})(n^{3}) = n^{11}$, what is $k$?
Multiplying with the same base means the exponents add: $k + 3 = 11$, so $k = 8$.
The ISEE puts $14$ (from $11 + 3$) and $33$ (from $11 \times 3$) in the choices for exactly this kind of question. Same base and multiplying? Add the exponents. That's it.
The power-of-a-power rule has its own trap, and it's all about the parentheses.
Simplify $\left(x^{4}\right)^{3}$.
That's three copies of $x^{4}$, which is $x^{4} \cdot x^{4} \cdot x^{4} = x^{12}$. So the exponents multiply: $\left(x^{4}\right)^{3} = x^{4 \cdot 3} = x^{12}$, not $x^{7}$.
Sometimes the same base is hiding. If one side has a plain number like $8$, rewrite it as a power of the base first.
If $2^{m} \cdot 8 = 2^{10}$, what is $m$?
The $8$ isn't a power of $2$ yet, so make it one: $8 = 2^{3}$. Now it's $2^{m} \cdot 2^{3} = 2^{10}$, and multiplying adds the exponents: $m + 3 = 10$, so $m = 7$.
Subtracting the $8$ itself ($10 - 8 = 2$) is the trap. Only exponents get added and subtracted here, never the plain number.
That's all three laws, plus the hidden base. Find the missing exponent.
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One more thing about stacked exponents before we move on, because the multiply rule only works with parentheses.
Watch out for exponents stacked with no parentheses. $\left(2^{3}\right)^{2} = 2^{6} = 64$, but $2^{3^{2}}$ means $2^{(3^{2})} = 2^{9} = 512$. Totally different numbers! The multiply rule only works when the parentheses are there.
Parentheses matter with negative numbers too. The exponent only grabs what sits right under it.
Exponents come before everything except parentheses. That's the E in PEMDAS from Section 0, and it's why the minus sign in $-3^{2}$ has to wait its turn.
Here's that picture written out, plus what happens with an odd power.
What are the values of $(-3)^{2}$ and $-3^{2}$?
In $(-3)^{2}$, the parentheses glue the minus sign to the $3$, so the base is $-3$: $(-3)^{2} = (-3)(-3) = 9$.
In $-3^{2}$, there are no parentheses, so the exponent only grabs the $3$. Square first, then put the minus sign on: $-3^{2} = -(3 \cdot 3) = -9$.
And a negative base to an odd power stays negative: $(-2)^{3} = (-2)(-2)(-2) = -8$.
Before you square anything, ask what the exponent grabbed. Some versions stack exponents too.
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If you played with the tiles, you already saw this coming. What's $\dfrac{b^{3}}{b^{3}}$? Every tile cancels, so it's $1$. But the division rule says it's $b^{3-3} = b^{0}$. So $b^{0}$ has to be $1$. And if there are more tiles on the bottom than the top, the leftovers live in the denominator. That's what a negative exponent is.
Here's the same idea as a pattern. Walk down the stairs one step at a time, dividing by the base each time, and don't stop at $b^{1}$.
Every step down was a division, and past $2^{0}$ the steps turned into fractions: one over a power. Flipping a number into "one over" it is the reciprocal from Section 0.
The reciprocal from Section 0: flip the fraction over. The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$, the reciprocal of $5$ is $\frac{1}{5}$, and a number times its reciprocal is always $1$.
A negative exponent indicates a reciprocal: $b^{-x}$ means "one divided by $b^{x}$."
Put that together with what the staircase showed about $b^{0}$, and you get the two rules for this part.
Zero/Negative Exponents (for $b \neq 0$):
$$b^{0}=1, \qquad b^{-x}=\dfrac{1}{b^{x}}$$
Why can't $b$ be $0$? Because $0^{-1}$ would mean $\dfrac{1}{0}$, and nobody gets to divide by zero.
Here's the rule in action, with the trap everyone falls into.
What is the value of $5^{-2}$?
$$5^{-2} = \frac{1}{5^{2}} = \frac{1}{25}$$
Your gut says the answer should be negative. Your gut is wrong. A negative exponent never makes a number negative. It flips it into a fraction. So $-25$ and $-10$ are both trap answers.
So where does a minus sign come from? Only from the base, and from where the parentheses put it. Sort these and see.
A zero exponent makes $1$, and a negative one makes a fraction, never a negative number. Try a few, some with the laws mixed in.
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Exponents also distribute over multiplication and division. Everything inside the parentheses gets the exponent.
Power on Products/Quotients (for $a, b \neq 0$):
$$(ab)^{x}=a^{x}b^{x}, \qquad \left(\dfrac{a}{b}\right)^{x}=\dfrac{a^{x}}{b^{x}}$$
Put the reciprocal idea together with a fraction, and you get the slickest shortcut in the section: a negative exponent on a fraction just flips the fraction.
Negative Exponent Rule (for $a, b \neq 0$):
$$\left(\dfrac{a}{b}\right)^{-x}=\left(\dfrac{b}{a}\right)^{x}$$
Example: $2^{-3}=\left(\dfrac{2}{1}\right)^{-3}=\left(\dfrac{1}{2}\right)^{3}=\dfrac{1^{3}}{2^{3}}=\dfrac{1}{8}$
Here it is in motion: the flip uses up the minus sign, and then the power goes to the top and the bottom.
Flip first, then power the top and the bottom.
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Now let's throw everything at one problem. Deal with the numbers and the letters separately, and they can't trip each other up.
Simplify and write with positive exponents: $\dfrac{(4m^{-3})(6m^{2})}{8m^{-2}}$
Numbers first: $\dfrac{4 \cdot 6}{8} = \dfrac{24}{8} = 3$.
Now the $m$'s. Add the exponents on top, then subtract the bottom one:
$$m^{-3+2-(-2)} = m^{1} = m$$
So the whole thing is $3m$. That double negative, $-(-2)$, is where most people go wrong.
Numbers with numbers, letters with letters. Some versions come as a rectangle, where the area is one side times the other.
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When the whole product sits inside parentheses with an exponent outside, the Powers of Products rule kicks in first.
Simplify and write with positive exponents: $\left(3a^{2}b^{-3}\right)^{2}\cdot b^{5}$
The outside $2$ hits everything in the parentheses, including the $3$:
$$\left(3a^{2}b^{-3}\right)^{2} = 3^{2}a^{4}b^{-6} = 9a^{4}b^{-6}$$
Then multiply by $b^{5}$: $9a^{4}b^{-6+5} = 9a^{4}b^{-1} = \dfrac{9a^{4}}{b}$.
That last step, moving $b^{-1}$ under the bar, has its own trap.
"Write with positive exponents" means slide anything with a negative exponent across the fraction bar and flip its sign. But only the thing wearing the negative exponent moves! $3x^{-2} = \dfrac{3}{x^{2}}$, not $\dfrac{1}{3x^{2}}$. The $3$ never had a negative exponent, so it stays put.
The number out front is the factor everyone forgets to raise. Make the outside exponent hit every single factor.
Now on paper: every factor gets the outside exponent.
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The ISEE also tests these rules with quantitative comparisons, the two-column questions from Section 1. Simplify each column first, then compare.
Assume $x > 0$.
Column A is $x^{3 + (-1)} = x^{2}$. Now try a few values. If $x = 2$, Column A is $4$ and Column B is $2$, so A wins. If $x = 1$, both are $1$. If $x = \tfrac{1}{2}$, Column A is $\tfrac{1}{4}$ and Column B is $\tfrac{1}{2}$, so now B wins. Different winners, so the answer is (D), it can't be determined.
The trap is thinking squaring always makes a number bigger. Fractions between $0$ and $1$ get smaller when you square them.
You'll get a question like that in the Mixed Practice at the bottom, where any idea from this section is fair game.
Two exponents show up so often that they get their own names. They're named after shapes, and there's a good reason for that.
The square of $a$ is $a^{2}=a\cdot a$ (read "a squared").
Every operation in math has an undo button. For squaring, it's the square root.
For $a \ge 0$, the square root $\sqrt{a}$ is the nonnegative number whose square is $a$.
Same story one dimension up.
The cube of $a$ is $a^{3}=a\cdot a\cdot a$ (read "a cubed").
And its undo button:
The cube root $\sqrt[3]{a}$ is the number (possibly negative) whose cube is $a$.
Notice the difference: a square root is never negative, but a cube root can be. $\sqrt[3]{-8} = -2$ because $(-2)^{3} = -8$. But $\sqrt{-4}$ isn't a real number at all, since no real number times itself is negative. Now here's why they're called squares and cubes.
A square with side $3$ has area $3^{2} = 9$, and going backwards, $\sqrt{9} = 3$ is the side of a square with area $9$. A cube with side $2$ has volume $2^{3} = 8$, so $\sqrt[3]{8} = 2$. Run a few roots backwards yourself, and watch for a root that can't be real.
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The numbers you can build exactly like this, as a whole square or a whole cube, get a special name.
A number is a perfect $n^{\text{th}}$ power if it equals $b^{n}$ for some integer $b$. A perfect $2^{\text{nd}}$ power is a perfect square, and a perfect $3^{\text{rd}}$ power is a perfect cube.
To test a number, check it against the squares and cubes on either side of it.
Is $49$ a perfect square? Is it a perfect cube?
$49 = 7^{2}$, so yes, it's a perfect square (a perfect $2^{\text{nd}}$ power). But $3^{3} = 27$ and $4^{3} = 64$, and $49$ falls in between, so it is not a perfect cube.
That check is only fast if you know the lists by heart.
Memorize these cold, because they're the key to every root problem. The perfect squares through $15^{2}$ are $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225$. The perfect cubes through $5^{3}$ are $1, 8, 27, 64, 125$, plus $10^{3} = 1000$. And notice that $64$ is on both lists: $64 = 8^{2} = 4^{3}$.
With the lists in your head, spotting the perfect one is quick.
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So what happens when the exponent is a fraction? That has a name too.
A rational exponent is an exponent that's a fraction.
It turns out a fraction exponent is a root and a power wearing a trench coat. The bottom of the fraction is the root, and the top is the power.
Radicals & Rational Exponents (for $b > 0$, where $x$ and $y$ are integers and $y > 0$):
$$b^{x/y}=\sqrt[y]{b^{\,x}}=\left(\sqrt[y]{b}\right)^{x}$$
Example: $16^{3/2}=\sqrt{16^{3}}=\left(\sqrt{16}\right)^{3}$. (With a positive exponent, $b = 0$ works too.)
Fraction exponents aren't weird new things. They fill in the gaps between the whole-number powers.
Both versions of the formula give the same answer, but one of them is way nicer to work with. Watch them race.
Here's the first race on paper.
Evaluate $16^{3/2}$.
Root first: $\sqrt{16} = 4$, then $4^{3} = 64$.
Power first works too, but look at it: $16^{3} = 4096$, and then you need $\sqrt{4096}$. Good luck doing that one in your head.
So the order matters for your sanity, even though it doesn't matter for the answer.
Always take the root first. The root makes the number small, and small numbers are easy to raise to a power. Power first makes the number huge, and huge numbers are miserable to take roots of.
When the fraction exponent is negative, handle the negative first by flipping, then do the root and power like usual.
Evaluate $\left(\tfrac{1}{8}\right)^{-2/3}$.
The negative flips the fraction: $\left(\tfrac{1}{8}\right)^{-2/3} = 8^{2/3}$. Now root first: $\sqrt[3]{8} = 2$, and $2^{2} = 4$.
Flip if it's negative, root, then power.
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The ISEE also loves asking you to translate between the two ways of writing the same thing.
Which is equivalent to $\sqrt[4]{x^{3}}$ (for $x \ge 0$)?
The root becomes the bottom of the fraction and the power becomes the top: $\sqrt[4]{x^{3}} = x^{3/4}$. The trap answer is $x^{4/3}$, which flips the fraction. Remember: the root lives in the basement.
Root in the basement, power upstairs. Translate these.
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The hardest version mixes the translation with a power of a power.
Find $a$ so that $\left(x^{3/4}\right)^{a} = \sqrt{x^{3}}$ (assume $x > 0$).
Rewrite the right side with an exponent: $\sqrt{x^{3}} = x^{3/2}$. The left side is a power of a power, so the exponents multiply:
$$\frac{3}{4} \cdot a = \frac{3}{2}, \quad\text{so}\quad a = \frac{3}{2} \cdot \frac{4}{3} = 2$$
Rewrite the root as an exponent, then divide.
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First, some names. The root sign itself has one.
The root sign $\sqrt{\phantom{x}}$ is called a radical, and so is any expression written with one.
And so does the number living under it.
In $\sqrt[n]{a}$, the number $a$ under the root sign is the radicand.
When the ISEE asks you to "simplify" a root, it has a very specific finish line in mind.
A root is in simplified form when the radicand has no perfect $n^{\text{th}}$-power factors (other than $1$), no fractions inside the root, and the denominator is rational (no roots in the denominator).
Whoever designed the radical sign gave it a roof that just... stops. How far does the roof reach? Better guess right, because $\sqrt{4} \cdot 9 = 18$ but $\sqrt{4 \cdot 9} = 6$.
Before we simplify anything, here's the rule that makes it possible. Roots get along with multiplying and dividing, but not with adding.
Roots of products and quotients (for $a \ge 0$ and $b > 0$):
$$\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}, \qquad \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$$
But a root never splits over a plus or minus sign: $\sqrt{9+16}=\sqrt{25}=5$, not $\sqrt{9}+\sqrt{16}=7$.
Why not? Look at the tiles. A $3$-by-$3$ square and a $4$-by-$4$ square don't make a $7$-by-$7$ square.
Those squares, $3^{2} + 4^{2} = 5^{2}$, are the Pythagorean theorem in disguise. In Section 6 they'll sit on the three sides of a right triangle.
Here are both halves of the rule in one example.
What are $\sqrt{2}\cdot\sqrt{18}$ and $\sqrt{36+64}$?
Multiply under one root: $\sqrt{2}\cdot\sqrt{18} = \sqrt{36} = 6$.
Add first, then take the root: $\sqrt{36+64} = \sqrt{100} = 10$. Splitting it into $6 + 8 = 14$ is the trap.
Before each reveal, call it: does the root split or not?
Times and divide split; plus and minus don't. Use the rule.
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Now the main event: simplifying a root. The game plan is to find the biggest perfect square hiding inside the number, and pull it out.
Can't spot the biggest perfect square? Use prime factorization from Section 0: $72 = 2^{3} \cdot 3^{2}$. Every matching pair of primes is a perfect square, so one from each pair comes out ($2 \cdot 3 = 6$), and the leftover $2$ stays inside: $6\sqrt{2}$.
The picture of $72$ as two $6$-by-$6$ squares, written out:
Simplify $\sqrt{72}$.
The biggest perfect square that goes into $72$ is $36$:
$$\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36}\cdot\sqrt{2} = 6\sqrt{2}$$
If you only spotted $4$, you'd get $2\sqrt{18}$. That's equal, but it's not simplified, because $18$ still has a $9$ hiding in it. The ISEE puts exactly that answer in the choices.
The prime factor way from the throwback, as a game: every matching pair is a perfect square, so one copy of it comes out.
Variables work the same way. An even exponent comes out of a square root cut in half, and an odd exponent leaves one copy behind.
Simplify $\sqrt{300x^{7}y^{4}z}$ (assume $x, y, z \ge 0$).
Go piece by piece. $300 = 100 \cdot 3$, so a $10$ comes out and a $3$ stays in. $x^{7} = x^{6}\cdot x$, so $x^{3}$ comes out and one $x$ stays in. $y^{4}$ comes out completely as $y^{2}$. And $z$ is stuck inside.
$$\sqrt{300x^{7}y^{4}z} = 10x^{3}y^{2}\sqrt{3xz}$$
And the same trick works for cube roots, with one change.
Cube roots work exactly the same, except you're hunting for groups of three. $\sqrt[3]{54x^{5}} = \sqrt[3]{27 \cdot 2 \cdot x^{3} \cdot x^{2}} = 3x\sqrt[3]{2x^{2}}$. Divide each exponent by $3$: the quotient comes out, and the remainder stays in.
Pull out the biggest square (or cube) and stop only when nothing else can come out.
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The last part of the definition says no roots in the denominator. If one shows up, multiply the top and bottom by that root to kick it out. That move has a name.
Rewriting a fraction so there's no root in its denominator is called rationalizing the denominator.
Here's the trick in motion: multiply by the root over itself, which is just $1$, and the root on the bottom disappears.
Here's another one on paper.
Simplify $\dfrac{12}{\sqrt{6}}$.
$$\frac{12}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{12\sqrt{6}}{6} = 2\sqrt{6}$$
Not every multiplier works. Pick one and see what it does to the bottom.
The middle part of the definition, no fractions inside the root, uses the root rule for fractions, and then the same trick.
Simplify $\sqrt{\dfrac{3}{4}}$ and $\sqrt{\dfrac{2}{3}}$.
Split each root: $\sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{\sqrt{4}} = \dfrac{\sqrt{3}}{2}$. Done, since the bottom came out a whole number.
$\sqrt{\dfrac{2}{3}} = \dfrac{\sqrt{2}}{\sqrt{3}}$ still has a root on the bottom, so multiply top and bottom by $\sqrt{3}$:
$$\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$$
Clear the bottom, and simplify what's left.
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Scientists deal with numbers like the distance to the Sun and the width of a bacterium, and nobody wants to count fifteen zeros. So they write every number as a "normal-sized" number times a power of ten. First, a name for the long way.
A number written out the usual way, with all its digits, is in standard form.
Scientific notation is the shorthand.
A number is in scientific notation when written $S\times 10^{k}$ with $1\le S<10$, where $k$ is an integer (positive or negative).
That $1 \le S < 10$ rule is the whole thing. $S$ needs exactly one nonzero digit in front of the decimal point. To get there, you slide the decimal point, and the exponent keeps track of the slides so the number doesn't actually change.
Every hop to the right makes $S$ ten times bigger, so the exponent has to drop by one to make up for it. Every hop left does the opposite. Tiny numbers end up with negative exponents and huge numbers end up with positive ones.
Write $0.00072$ in scientific notation.
Hop the decimal point right until there's one nonzero digit in front: $0.00072 \to 7.2$. That's $4$ hops right, so the exponent is $-4$:
$$0.00072 = 7.2\times 10^{-4}$$
Watch for $72\times 10^{-5}$ in the answer choices. It's the same number, but $72$ isn't between $1$ and $10$, so it's not scientific notation.
Count the hops, and let the direction pick the sign.
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Going back the other way is the same hops, run in reverse: the exponent tells you how many places to move the point, and which way.
Write $5.6\times 10^{-4}$ in standard form.
A negative exponent means a small number, so hop the point $4$ places left, filling in zeros as you go:
$$5.6\times 10^{-4} = 0.00056$$
Count the hops, not the zeros. $0.0056$ is only $3$ hops, and $56{,}000$ hopped the wrong way.
Now unpack a few back into standard form.
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Multiplying and dividing are the easy ones: deal with the $S$'s, deal with the powers of ten, then clean up.
Compute $(2.5\times 10^{4})(6\times 10^{-7})$.
Multiply the fronts and add the exponents: $2.5 \times 6 = 15$ and $10^{4} \cdot 10^{-7} = 10^{-3}$. That gives $15\times 10^{-3}$.
Don't stop there! $15$ is too big to be $S$. Hop the decimal one place left and bump the exponent up one:
$$15\times 10^{-3} = 1.5\times 10^{-2}$$
That cleanup step is where points get lost, so here it is slowed down, both ways.
Dividing works the same way, with the exponents subtracting.
Compute $\dfrac{8.4\times 10^{-2}}{2.1\times 10^{5}}$.
Divide the fronts and subtract the exponents: $\dfrac{8.4}{2.1} = 4$ and $10^{-2-5} = 10^{-7}$. So the answer is $4\times 10^{-7}$.
Fronts with fronts, powers with powers, then tidy up the front.
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Adding and subtracting are where the traps live, because you can't just add the exponents anymore.
Compute $(6.3\times 10^{4}) + (5.2\times 10^{3})$.
The powers of ten don't match, so first rewrite the smaller number with $10^{4}$: $5.2\times 10^{3} = 0.52\times 10^{4}$. Now they're like terms:
$$6.3\times 10^{4} + 0.52\times 10^{4} = 6.82\times 10^{4}$$
Line them up yourself: change the second number's power until it matches, and watch the digits stay in their columns.
That matching step is the one difference between these and the multiplying problems.
Multiplying or dividing? The exponents can be anything. Adding or subtracting? The exponents must match first, just like you need a common denominator to add fractions. Anyone who writes $11.5\times 10^{7}$ for the example above is about to lose a point.
Match the powers, then combine the fronts.
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Finally, the word problem version. These are usually about really tiny (or really huge) things, so here's a tour of the small end of the scale first.
Every tick on that line is a power of ten, so comparing sizes means comparing exponents first. Put these in order.
These problems almost always involve a unit conversion, so know that $1$ cm $= 10^{-2}$ m and $1$ mm $= 10^{-3}$ m.
A grain of pollen is about $2.5\times 10^{-5}$ meters wide. How many grains of pollen placed end to end would span $1$ centimeter?
Convert first: $1$ cm $= 10^{-2}$ m. Then divide the distance by the size of one grain:
$$\frac{10^{-2}}{2.5\times 10^{-5}} = \frac{1}{2.5}\times 10^{-2-(-5)} = 0.4\times 10^{3} = 4\times 10^{2}$$
So it takes about $400$ grains. If you forgot to convert centimeters to meters, your answer would be $100$ times too big.
Here are those $400$ grains, lined up a hundred at a time.
Convert the units, then divide one size by the other.
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One last place exponents hide: in anything that keeps multiplying. A colony of bacteria that doubles every hour isn't adding a fixed number each hour, it's multiplying by $2$ each hour, and that's an exponent in disguise.
Exponential growth (or exponential decay) is change by the same factor every period: the amount is multiplied by that factor once per period, so it grows (or shrinks) faster and faster (or slower and slower).
Count the periods, and the exponent is that count:
Doubling has factor $2$, halving has factor $\frac{1}{2}$, and growing by $10\%$ has factor $1.1$. The exponent counts periods, not hours.
A colony of bacteria doubles every $3$ hours. If there are $50$ bacteria now, how many are there after $12$ hours?
In $12$ hours there are $12 \div 3 = 4$ doublings, so $50 \cdot 2^4 = 50 \cdot 16 = 800$.
The trap is $50 \cdot 2 \cdot 4 = 400$: that adds $50$ every period instead of doubling what's there. And $50 \cdot 2^{12}$ would be doubling every hour.
Percents work the same way, once you turn the percent into a factor. Growing by $10\%$ means keeping $110\%$, so the factor is $1.1$.
A town of $1{,}000$ people grows by $10\%$ each year. What is the population after $3$ years?
$1{,}000 \cdot 1.1^3 = 1{,}000 \cdot 1.331 = 1{,}331$.
Adding the percents, $3 \cdot 10\% = 30\%$, gives $1{,}300$. That's the same trap from Section 1: each year's $10\%$ is taken of a new, bigger population.
This is Percents in a Row with the same percent every time: multiply the factors, and the factors are all equal, so it's a power.
Halving never reaches zero. Half of $800$ is $400$, then $200$, then $100$: after $n$ halvings there is always $\frac{800}{2^n}$ left. If a decay question's answer choices include $0$, that's the trap.
Count the periods first, then raise the factor to that power:
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Every question so far came right after its idea, which is a hint the real test never gives. Down here the hint is gone. First, the quantitative comparison: two columns, four fixed choices, and any idea from this section fair game.
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Next, the mixed set. Every question type on this page except the quantitative comparison goes into one hat, and each click draws a new one at random. Flip the switch to type your answer: fractions work (3/2), so do roots (2sqrt3 or sqrt6/3, in simplest form) and scientific notation (1.28 x 10^3). A few types need their answer choices, and they'll say so.
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Alright, time for something with a little heat. These equations have the variable stuck up in the exponent. The trick is to rewrite both sides as powers of the same base (like turning $27$ into $3^{3}$ and $9$ into $3^{2}$), just like the hidden $8 = 2^{3}$ from earlier. Once the bases match, the exponents have to match too, and you're left with a plain old equation. Take your time with the parentheses.
Solve for $x$: $4^{x} = 8^{x-1}$.
Both $4$ and $8$ are powers of $2$: $4 = 2^{2}$ and $8 = 2^{3}$. Rewrite both sides, then use the power-of-a-power rule:
$$\left(2^{2}\right)^{x} = \left(2^{3}\right)^{x-1}, \quad\text{so}\quad 2^{2x} = 2^{3x-3}$$
The bases match, so the exponents match: $2x = 3x - 3$, so $x = 3$. Check it: $4^{3} = 64$ and $8^{2} = 64$.
Your turn. Same base first, then set the exponents equal.
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