Powers & Radicals · Lecture 08 · alg-powers-radicals
📺 Course materials — Lecture 08:
lecture video · handout PDF (Lecture 8) · board notes (08.jpg) · homework answer key (Chapter 8)
— all in the
amc8/video folder.
·
🖨 Printable PDF (answers & solutions at the back)
🔑 Key facts
A power $a^n$ is $n$ copies of $a$ multiplied together, and two laws run the whole chapter:
$a^m\cdot a^n=a^{m+n}$ and $(a^m)^n=a^{mn}$.
A radical $\sqrt{x}$ is the nonnegative number whose square is $x$ — powers and roots undo each other.
Nearly every AMC 8 problem here is won by rewriting everything with the same base or the same exponent.
🧰 Instructor's toolbox (from the lecture video & board notes)
2 · The two exponent laws that do everything.
$a^m\cdot a^n=a^{m+n}$ (same base: add exponents) and $(a^m)^n=a^{mn}$ (power of a power: multiply).
That's how $100^{2x}$ becomes $10^{4x}$ in Practice 4, how $2^{18}=(2^6)^3=64^3$ turns a huge range into
a cube count in Practice 7, and how every comparison in Example 3 is set up.
3 · Isolate the power first, then match bases. Before any exponent trick,
get the power alone on one side: $3^p+3^4=90 \Rightarrow 3^p=90-81=9=3^2$, so $p=2$ — then read the exponent off
matching bases (Example 1). Same move three times: $2^r=76-44=32=2^5$, $6^s=1421-125=1296=6^4$.
4 · Prime factorization reads off exponents. Factorizations are unique,
so if $2^w\cdot3^x\cdot5^y\cdot7^z=588=2^2\cdot3\cdot7^2$, then $w,x,y,z$ are forced: $2,1,0,2$ —
a missing prime means exponent $0$, not "no answer" (Example 2). Factor trees are the door into every
"what power of … divides …" question.
5 · Compare big powers with a common exponent. You can't eyeball
$10^8$ vs $5^{12}$ vs $2^{24}$ — so give them the same exponent, $\gcd(8,12,24)=4$:
\[10^8=(10^2)^4=100^4,\qquad 5^{12}=(5^3)^4=125^4,\qquad 2^{24}=(2^6)^4=64^4,\]
and now just compare the bases $64<100<125$ (Example 3). Same exponent → compare bases;
same base → compare exponents.
6 · $\sqrt{9}=3$, never $\pm3$. The radical sign means the principal
(nonnegative) root — the board underlines "$\sqrt9 = 3$ or $-3$" as the classic error. But the
equation $x^2=9$ really does have two roots, $x=\pm3$ — that's where the $\pm$ lives.
Consequence: $\sqrt{a^2}=|a|$, so $\sqrt{(3-2\sqrt3)^2}=|3-2\sqrt3|=2\sqrt3-3$, because
$2\sqrt3\approx3.46>3$ (Practice 5).
7 · Radical survival kit.
$\sqrt{A\cdot B}=\sqrt A\cdot\sqrt B$ and $\sqrt{\tfrac AB}=\tfrac{\sqrt A}{\sqrt B}$ (for $A,B\ge0$, $B\ne0$).
Simplify by pulling out the perfect square factor: $\sqrt{40}=\sqrt{4\times10}=2\sqrt{10}$.
Clear a root from a denominator by multiplying by it:
$\sqrt{\tfrac23}=\tfrac{\sqrt2}{\sqrt3}\cdot\tfrac{\sqrt3}{\sqrt3}=\tfrac{\sqrt6}{3}$ — both straight from the board.
8 · Nested radicals unwind from the inside. For $\sqrt{16\sqrt{8\sqrt4}}$,
start at the deepest layer and work out: $\sqrt4=2$, then $\sqrt{8\cdot2}=\sqrt{16}=4$, then
$\sqrt{16\cdot4}=\sqrt{64}=8$ (Example 4). Never try to distribute the outer root across the inside.
9 · Conjugates and $a^2-b^2=(a+b)(a-b)$. The difference of squares runs both
directions here: it factors monsters like $13^4-11^4=(13^2+11^2)(13^2-11^2)$ (Practice 6) and
$k^2-1=(k-1)(k+1)$ (Fresh 10), and it collapses radical conjugates:
$(9-6\sqrt2)(9+6\sqrt2)=81-72=9$. When a sum of ugly radicals is asked for, square the whole
expression, simplify, then take the root back (Challenge C2).
10 · Estimate: round, then shift decimal points in pairs.
$0.000315\times7{,}928{,}564\approx0.0003\times8{,}000{,}000$ — now move one factor up by $1000$ and the other
down by $1000$: $0.3\times8000=2400$ (Practice 1). Multiplying and dividing by the same power of $10$
keeps the product honest and the zeros manageable.
📖 Worked examples
2013 AMC 8 #15 · match the base
Example 1. If $3^p+3^4=90$, $2^r+44=76$, and $5^3+6^s=1421$, what is the product of $p$, $r$, and $s$?
(A) 27(B) 40(C) 50(D) 70(E) 90
Key idea: isolate each power, then write both sides with the same base.
$3^p=90-81=9=3^2 \Rightarrow p=2$.
$2^r=76-44=32=2^5 \Rightarrow r=5$.
$6^s=1421-125=1296=6^4 \Rightarrow s=4$.
Product: $2\cdot5\cdot4=40$.
Answer: B
2011 AMC 8 #17 · prime factorization
Example 2. Let $w$, $x$, $y$, and $z$ be whole numbers. If $2^w\cdot3^x\cdot5^y\cdot7^z=588$, then what does $2w+3x+5y+7z$ equal?
(A) 21(B) 25(C) 27(D) 35(E) 56
Key idea: factor $588$ into primes: $588=4\cdot147=2^2\cdot3\cdot7^2$.
Prime factorization is unique, so $w=2$, $x=1$, $y=0$ (no factor of $5$!), $z=2$.
Then $2w+3x+5y+7z=4+3+0+14=21$.
Answer: A
2010 AMC 8 #24 · comparing powers
Example 3. What is the correct ordering of the three numbers $10^8$, $5^{12}$, and $2^{24}$?
(A) $2^{24}<10^8<5^{12}$(B) $2^{24}<5^{12}<10^8$(C) $5^{12}<2^{24}<10^8$(D) $10^8<5^{12}<2^{24}$(E) $10^8<2^{24}<5^{12}$
Key idea: the exponents $8, 12, 24$ share the factor $4$ — rewrite everything as a fourth power:
$10^8=(10^2)^4=100^4$, $\quad 5^{12}=(5^3)^4=125^4$, $\quad 2^{24}=(2^6)^4=64^4$.
Same exponent, so compare bases: $64<100<125$, giving $2^{24}<10^8<5^{12}$.
Answer: A
2017 AMC 8 #3 · nested radicals
Example 4. What is the value of the expression $\sqrt{16\sqrt{8\sqrt{4}}}$ ?
(A) 4(B) $4\sqrt{2}$(C) 8(D) $8\sqrt{2}$(E) 16
Key idea: unwind from the deepest radical:
$\sqrt4=2$, so the middle layer is $\sqrt{8\cdot2}=\sqrt{16}=4$, and the outside is
$\sqrt{16\cdot4}=\sqrt{64}=8$.
Answer: C
✏️ Practice
Try each one before opening the solution.
2017 AMC 8 #4
1. When $0.000315$ is multiplied by $7{,}928{,}564$ the product is closest to which of the following?
(A) 210(B) 240(C) 2100(D) 2400(E) 24000
Solution
Round: $0.0003\times8{,}000{,}000$. Shift by $1000$ both ways: $0.3\times8000=2400$.
Answer: D
2022 AMC 8 #2
2. Consider these two operations: \[a\,\blacklozenge\,b=a^2-b^2 \qquad\qquad a\,\bigstar\,b=(a-b)^2\]
What is the value of $(5\,\blacklozenge\,3)\,\bigstar\,6$?
(A) $-20$(B) 4(C) 16(D) 100(E) 220
Solution
Inside first: $5\,\blacklozenge\,3=5^2-3^2=25-9=16$. Then $16\,\bigstar\,6=(16-6)^2=10^2=100$.
($a^2-b^2$ and $(a-b)^2$ are
not the same thing — that's the whole point of the problem.)
Answer: D
2016 AMC 8 #7
3. Which of the following numbers is not a perfect square?
(A) $1^{2016}$(B) $2^{2017}$(C) $3^{2018}$(D) $4^{2019}$(E) $5^{2020}$
Solution
A power is a perfect square when it can be written as $(\text{something})^2$ — i.e. its exponent is even
(over a prime base). Even exponents: A, C, E ✓. And $4^{2019}=(2^2)^{2019}=2^{4038}$ — even too ✓.
Only $2^{2017}$ has an odd exponent of a prime.
Answer: B
2016 AMC 10A #2
4. For what value of $x$ does $10^x\cdot100^{2x}=1000^5$?
(A) 1(B) 2(C) 3(D) 4(E) 5
Solution
Everything is a power of $10$:
$10^x\cdot(10^2)^{2x}=10^x\cdot10^{4x}=10^{5x}$ and $1000^5=(10^3)^5=10^{15}$.
So $5x=15$, $x=3$.
Answer: C
2021 AMC 10B #2
5. What is the value of $\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}$?
(A) 0(B) $4\sqrt{3}-6$(C) 6(D) $4\sqrt{3}$(E) $4\sqrt{3}+6$
Solution
$\sqrt{a^2}=|a|$, so the expression is $|3-2\sqrt3|+|3+2\sqrt3|$.
Since $2\sqrt3\approx3.46>3$, the first term is $2\sqrt3-3$ (not $3-2\sqrt3$!), the second is $3+2\sqrt3$.
Sum: $(2\sqrt3-3)+(2\sqrt3+3)=4\sqrt3$.
Answer: D
2016 AMC 8 #15
6. What is the largest power of $2$ that is a divisor of $13^4-11^4$?
(A) 8(B) 16(C) 32(D) 64(E) 128
Solution
Factor twice: $13^4-11^4=(13^2+11^2)(13^2-11^2)=(169+121)(169-121)=290\cdot48$.
Count the $2$s: $290=2\cdot145$ and $48=2^4\cdot3$, so the product has $2^5=32$.
Answer: C
2018 AMC 8 #25
7. How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?
(A) 4(B) 9(C) 10(D) 57(E) 58
Solution
$2^8+1=257$: since $6^3=216<257\le343=7^3$, the smallest cube in range is $7^3$.
$2^{18}=(2^6)^3=64^3$, and $64^3\le2^{18}+1$, so the largest is $64^3$.
Cubes of $7,8,\ldots,64$: that's $64-7+1=58$.
Answer: E
🆕 Fresh from the latest exams
The same ideas, exactly as they appeared on the most recent tests.
2024 AMC 8 #4
8. When Yunji added all the integers from $1$ to $9$, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
(A) 5(B) 6(C) 7(D) 8(E) 9
Solution
$1+2+\cdots+9=45$. Leaving out a number from $1$ to $9$ puts the sum between $36$ and $44$ —
the only perfect square there is $36$, so she left out $45-36=9$.
Answer: E
2026 AMC 8 #9
9. What is the value of this expression? \[\frac{\sqrt{16\sqrt{81}}}{\sqrt{81\sqrt{16}}}\]
(A) $\frac{4}{9}$(B) $\frac{2}{3}$(C) 1(D) $\frac{3}{2}$(E) $\frac{9}{4}$
Solution
Inside out, top and bottom:
numerator $\sqrt{16\cdot9}=\sqrt{144}=12$; denominator $\sqrt{81\cdot4}=\sqrt{324}=18$.
So the value is $\dfrac{12}{18}=\dfrac23$.
Answer: B
2025 AMC 8 #23
10. How many four-digit numbers have all three of the following properties? (I) The tens and ones digit are both $9$. (II) The number is $1$ less than a perfect square. (III) The number is the product of exactly two prime numbers.
(A) 0(B) 1(C) 2(D) 3(E) 4
Solution
Write the number as $k^2-1=(k-1)(k+1)$. It ends in $99$, so $k^2$ ends in $00$ — $k$ is a multiple of $10$,
and four digits force $k\in\{40,50,\ldots,100\}$.
For exactly two prime factors we need $k-1$ and $k+1$
both prime:
$39,49,69,81,91,99$ all fail; only $k=60$ works ($59$ and $61$ are twin primes), giving $3599=59\cdot61$.
Exactly one such number.
Answer: B
⭐ Challenge (AMC 10 level)
2003 AMC 10B #9 · exponent equation
C1. Find the value of $x$ that satisfies the equation \[25^{-2}=\frac{5^{48/x}}{5^{26/x}\cdot25^{17/x}}.\]
(A) 2(B) 3(C) 5(D) 6(E) 9
Solution
Everything in base $5$. Left: $25^{-2}=5^{-4}$. Right: $25^{17/x}=5^{34/x}$, so the exponent is
$\dfrac{48}{x}-\dfrac{26}{x}-\dfrac{34}{x}=\dfrac{-12}{x}$.
Match exponents: $-4=\dfrac{-12}{x} \Rightarrow x=3$.
Answer: B
course problem · radical conjugates
C2. Which of the following is equal to $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$?
(A) $3\sqrt{2}$(B) $2\sqrt{6}$(C) $\frac{7\sqrt{2}}{2}$(D) $3\sqrt{3}$(E) 6
Solution
Call the whole thing $S$ (note $S>0$) and square it:
\[S^2=(9-6\sqrt2)+(9+6\sqrt2)+2\sqrt{(9-6\sqrt2)(9+6\sqrt2)}=18+2\sqrt{81-72}=18+2\cdot3=24.\]
So $S=\sqrt{24}=2\sqrt6$.
Answer: B
🗝 Answer key
| Example | 1 | 2 | 3 | 4 |
|---|
| B | A | A | C |
|---|
| Practice | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|
| D | D | B | C | D | C | E | E | B | B |
|---|
| Challenge | C1 | C2 |
|---|
| B | B |
|---|
Sources: prep-course Lecture 08 (video + board notes + 9 worked examples, translated to English) and official AMC 8 problems from the local bank
(kb/content/banks/), which also supplied the extra on-topic practice (2016 #7, 2016 #15, 2018 #25, 2022 #2). Answers cross-checked between the course solutions and the bank's official answer keys.