Stay ahead of learning milestones! Enroll in a class over the summer!

Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
9 Will I make JMO?
EaZ_Shadow   17
N 8 minutes ago by sadas123
will I be able to make it... will the cutoffs will be pre-2024
17 replies
EaZ_Shadow
Feb 7, 2025
sadas123
8 minutes ago
9 ARML Location
deduck   40
N an hour ago by zhenghua
UNR -> Nevada
St Anselm -> New Hampshire
PSU -> Pennsylvania
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
40 replies
deduck
May 6, 2025
zhenghua
an hour ago
Preparing for Putnam level entrance examinations
Cats_on_a_computer   4
N 2 hours ago by Cats_on_a_computer
Non American high schooler in the equivalent of grade 12 here. Where I live, two the best undergraduates program in the country accepts students based on a common entrance exam. The first half of the exam is “screening”, with 4 options being presented per question, each of which one has to assign a True or False. This first half is about the difficulty of an average AIME, or JEE Adv paper, and it is a requirement for any candidate to achieve at least 24/40 on this half for the examiners to even consider grading the second part. The second part consists of long form questions, and I have, no joke, seen them literally rip off, verbatim, Putnam A6s. Some of the problems are generally standard textbook problems in certain undergrad courses but obviously that doesn’t translate it to being doable for high school students. I’ve effectively got to prepare for a slightly nerfed Putnam, if you will, and so I’ve been looking for resources (not just problems) for Putnam level questions. Does anyone have any suggestions?
4 replies
Cats_on_a_computer
Yesterday at 8:32 AM
Cats_on_a_computer
2 hours ago
Past USAMO Medals
sdpandit   1
N 3 hours ago by CatCatHead
Does anyone know where to find lists of USAMO medalists from past years? I can find the 2025 list on their website, but they don't seem to keep lists from previous years and I can't find it anywhere else. Thanks!
1 reply
sdpandit
Thursday at 7:44 PM
CatCatHead
3 hours ago
Marginal Profit
NC4723   1
N 5 hours ago by Juno_34
Please help me solve this
1 reply
NC4723
Dec 11, 2015
Juno_34
5 hours ago
An FE. Who woulda thunk it?
nikenissan   116
N Today at 9:12 AM by anudeep
Source: 2021 USAJMO Problem 1
Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]
116 replies
nikenissan
Apr 15, 2021
anudeep
Today at 9:12 AM
Romania NMO 2023 Grade 11 P1
DanDumitrescu   15
N Today at 5:46 AM by anudeep
Source: Romania National Olympiad 2023
Determine twice differentiable functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which verify relation

\[
    \left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
    \]
15 replies
DanDumitrescu
Apr 14, 2023
anudeep
Today at 5:46 AM
Subset Ordered Pairs of {1, 2, ..., 10}
ahaanomegas   11
N Today at 5:27 AM by cappucher
Source: Putnam 1990 A6
If $X$ is a finite set, let $X$ denote the number of elements in $X$. Call an ordered pair $(S,T)$ of subsets of $ \{ 1, 2, \cdots, n \} $ $ \emph {admissible} $ if $ s > |T| $ for each $ s \in S $, and $ t > |S| $ for each $ t \in T $. How many admissible ordered pairs of subsets $ \{ 1, 2, \cdots, 10 \} $ are there? Prove your answer.
11 replies
ahaanomegas
Jul 12, 2013
cappucher
Today at 5:27 AM
Putnam 2000 B4
ahaanomegas   6
N Today at 1:53 AM by mqoi_KOLA
Let $f(x)$ be a continuous function such that $f(2x^2-1)=2xf(x)$ for all $x$. Show that $f(x)=0$ for $-1\le x \le 1$.
6 replies
ahaanomegas
Sep 6, 2011
mqoi_KOLA
Today at 1:53 AM
Another integral limit
RobertRogo   1
N Yesterday at 8:37 PM by alexheinis
Source: "Traian Lalescu" student contest 2025, Section A, Problem 3
Let $f \colon [0, \infty) \to \mathbb{R}$ be a function differentiable at 0 with $f(0) = 0$. Find
$$\lim_{n \to \infty} \frac{1}{n} \int_{2^n}^{2^{n+1}} f\left(\frac{\ln x}{x}\right) dx$$
1 reply
RobertRogo
Yesterday at 2:28 PM
alexheinis
Yesterday at 8:37 PM
AB=BA if A-nilpotent
KevinDB17   3
N Yesterday at 7:51 PM by loup blanc
Let A,B 2 complex n*n matrices such that AB+I=A+B+BA
If A is nilpotent prove that AB=BA
3 replies
KevinDB17
Mar 30, 2025
loup blanc
Yesterday at 7:51 PM
Very nice equivalence in matrix equations
RobertRogo   3
N Yesterday at 5:45 PM by Etkan
Source: "Traian Lalescu" student contest 2025, Section A, Problem 4
Let $A, B \in \mathcal{M}_n(\mathbb{C})$ Show that the following statements are equivalent:

i) For every $C \in \mathcal{M}_n(\mathbb{C})$ there exist $X, Y \in \mathcal{M}_n(\mathbb{C})$ such that $AX + YB = C$
ii) For every $C \in \mathcal{M}_n(\mathbb{C})$ there exist $U, V \in \mathcal{M}_n(\mathbb{C})$ such that $A^2 U + V B^2 = C$

3 replies
RobertRogo
Yesterday at 2:34 PM
Etkan
Yesterday at 5:45 PM
Miklos Schweitzer 1971_5
ehsan2004   2
N Yesterday at 5:25 PM by pi_quadrat_sechstel
Let $ \lambda_1 \leq \lambda_2 \leq...$ be a positive sequence and let $ K$ be a constant such that \[  \sum_{k=1}^{n-1} \lambda^2_k < K \lambda^2_n \;(n=1,2,...).\] Prove that there exists a constant $ K'$ such that \[  \sum_{k=1}^{n-1} \lambda_k < K' \lambda_n \;(n=1,2,...).\]

L. Leindler
2 replies
ehsan2004
Oct 29, 2008
pi_quadrat_sechstel
Yesterday at 5:25 PM
Cute matrix equation
RobertRogo   1
N Yesterday at 4:46 PM by loup blanc
Source: "Traian Lalescu" student contest 2025, Section A, Problem 2
Find all matrices $A \in \mathcal{M}_n(\mathbb{Z})$ such that $$2025A^{2025}=A^{2024}+A^{2023}+\ldots+A$$
1 reply
RobertRogo
Yesterday at 2:23 PM
loup blanc
Yesterday at 4:46 PM
EZ logarithm V2
fruitmonster97   42
N Apr 27, 2025 by cinnamon_e
Source: 2024 AIME I Problem 2
Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$?
42 replies
fruitmonster97
Feb 2, 2024
cinnamon_e
Apr 27, 2025
EZ logarithm V2
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 AIME I Problem 2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
fruitmonster97
2492 posts
#1 • 1 Y
Y by MathFan335
Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Bluesoul
898 posts
#2 • 1 Y
Y by sharknavy75
4xy=100, xy=25
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
balllightning37
389 posts
#3 • 1 Y
Y by Neeka_s08
I got 025. Anyone confirm?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mitsuihisashi14
121 posts
#4
Y by
Anyone confirm 025?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
gladIasked
648 posts
#5 • 1 Y
Y by isache
025, anyone? you could just solve for xy directly.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EvanZ
187 posts
#6 • 2 Y
Y by fura3334, dbnl
i trolled this for 2 hours and ended up guessing 25...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megarnie
5606 posts
#8
Y by
Let $y = x^k$. We want to find $x^{k+1}$.

We have $10 = \log_x(y^x) = xk$ and $10 =\log_y(x^{4y} ) = \frac{4y}{k} = \frac{4x^k}{k}$.

Multiplying equations gives $4x^{k+1} = 10 \cdot 10$, so answer is $\boxed{025}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cinnamon_e
703 posts
#9 • 5 Y
Y by MathFan335, Jack_w, dbnl, lucifer9838, Math_DM
this has to be the easiest aime problem ever

just multiply to get $x\log_xy\cdot 4y\log_yx=4xy=100\implies xy=\boxed{025}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
795 posts
#10
Y by
Our answer is
\[xy = \frac 14 \cdot 4xy = \frac 14 (x \log_xy)(4y \log_yx) = = \frac 14 \cdot 10 \cdot 10 = \boxed{025}. \quad \blacksquare\]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
williamxiao
2515 posts
#11
Y by
By log rules, we have $x\log_{x}y = 4y\log_{y}x = 10$. Letting $\log_{x}y = z$, we have $xz = \frac{4y}{z}= 10$. We get $z = \frac{10}{x} = \frac{4y}{10}$, so $4xy=100$ and $xy=025$
This post has been edited 1 time. Last edited by williamxiao, Feb 2, 2024, 6:04 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bluelinfish
1449 posts
#12
Y by
Simplify to get that $x\log_x(y) = 4y\log_y(x) = 10$. Therefore $100 = 4xy \log_x(y)\log_y(x) = 4xy$, implying that $xy = \boxed{025}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cappucher
96 posts
#13
Y by
I was worried on how easy this was, but $025$ is right...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MC413551
2228 posts
#14 • 1 Y
Y by matchaisgreen
x^10=y^x
y^10=x^4y
take 10th root of first equatoin to get
x=y^(x/10)
substitute into second equation
y^10=y^4xy/10
10=4xy/10
100=4xy
xy=25
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
gicyuraok2
1059 posts
#15
Y by
bro last year i guessed 25 on all my unanswered questions, and i was going to do the same thing this year until i solved this problem first. wut
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
exp-ipi-1
1074 posts
#16
Y by
MC413551 wrote:
x^10=y^x
y^10=x^4y
take 10th root of first equatoin to get
x=y^(x/10)
substitute into second equation
y^10=y^4xy/10
10=4xy/10
100=4xy
xy=25

yeah thats what i did
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
brainfertilzer
1831 posts
#17
Y by
??? $x\log_x y =10, 4y\log_y x = 10\implies 4xy(\log_xy)(\log_y x) = 100\implies 4xy = 100\implies xy = \boxed{025}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
plang2008
337 posts
#18
Y by
I can't be the only one who got the answer without even meaning to right?

Rewrite without logs as $x^{10} = y^x$ and $y^{10} = x^{4y}$. Raise the second equation to the $2.5$th power to get $y^{25} = x^{10y} = y^{xy}$ so $xy = \boxed{25}$.

After I got $xy = 25$ I attempted to find $x$ and $y$ before realizing what it was asking for...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mebookreader
1103 posts
#19
Y by
Bruh I didn't know how to do this.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zhu1223
2 posts
#20
Y by
I agree this should be 25
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Hayabusa1
478 posts
#21
Y by
$\log_x y=\frac{10}{x}\implies 4y\log_y x=4y\cdot \frac{x}{10}=10$. Thus, $xy=\frac{10\cdot 10}{4}=\boxed{025}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11320 posts
#22 • 2 Y
Y by FelipeGigena, MrDanny
fruitmonster97 wrote:
Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$?

Using $\log$ rules:
$$x\log_xy=4y\log_yx=10.$$Noticing the $\log_xy$ and $\log_yx$, recalling the identity $\log_xy\cdot\log_yx=1$ we multiply $x\log_xy$ and $4y\log_yx$ to get:
$$4xy=x\log_xy\cdot4y\log_yx=10\cdot10=100.$$Then $xy=\boxed{025}$.
This post has been edited 1 time. Last edited by jasperE3, Feb 3, 2024, 12:37 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Arrowhead575
2281 posts
#23
Y by
Most straightforward AIME question of all time?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
exp-ipi-1
1074 posts
#24
Y by
Arrowhead575 wrote:
Most straightforward AIME question of all time?

nah 2023 AIME 2 P1
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathboy282
2989 posts
#25
Y by
fruitmonster97 wrote:
Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$?

By Change of Base notice that log_x(y)=1/log_y(x). Thus, we have xlog_x(y)=(4y)log_y(x)=10. Then we get xlog_x(y)=4y/log_x(y)=10. Solving for log_x(y) we get log_x(y)=10/x. Then: 4y/(10/x)=10 => 2xy/5=10 => xy=25.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
orangesyrup
130 posts
#26
Y by
the ans is 025
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
xHypotenuse
778 posts
#27
Y by
this shouldve been problem 1
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
am07
316 posts
#28
Y by
Easiest problem on the test
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1005 posts
#29
Y by
When in doubt, spam log properties.
Observe that $\log_x(y^x) = x \frac{\log y}{\log x} = 10, \log_y(x^{4y}) = 4y\frac{\log x}{\log y} = 10$. Multiplying these two, we get $4xy = 100 \implies xy = \boxed{025}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sharknavy75
701 posts
#30
Y by
I did a very stupid solution.

I did the following
x^10 = y^x
y^10 = x^4y
multiply the 2 equations and get
x^10*y^10 = y^x*x^4y
1 = y^(x-10)x^(4y-10)
x-10 = 0 and 4y-10=0
x = 10 and y = 2.5
xy = 25
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megahertz13
3183 posts
#32
Y by
Video Solution & More
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Soumya_cena
17 posts
#33
Y by
fruitmonster97 wrote:
Real numbers $x$ and $y$ with $x,y>1$ satisfy $\log_x(y^x)=\log_y(x^{4y})=10.$ What is the value of $xy$?

Quite easy with logarithmic properties
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mr.Sharkman
500 posts
#34 • 1 Y
Y by blueprimes
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aidan0626
1901 posts
#35 • 1 Y
Y by Math_DM
pov: you get stuck on this problem in test because you didn't notice the "=10" part until like 10 minutes after
don't remember how i solved this in contest, i didn't notice you can just multiply them lol
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
A04572
9 posts
#36
Y by
xlogx y=4ylogyx=10 logxy*logyx=1, 4xy=100 xy=25 Can we solve for x and y directly?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ProMaskedVictor
43 posts
#37
Y by
$\log_x (y^x) =x\log_x y=10  -(i)$
$\log_y (x^{4y}) =4y\log_y x=10  -(ii)$
So, $(i) \times (ii): $
$4xy=100 \implies \boxed{xy=25}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rio_24
2 posts
#38
Y by
The Answer xy = 25
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
meduh6849
354 posts
#39
Y by
Yet another 025... once again, reinforcing its status as the most common AIME answer.
https://artofproblemsolving.com/community/c5h2018116
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
gracemoon124
872 posts
#40
Y by
storage
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
PYM7
5 posts
#41
Y by
Let $y = x^{k}$, where $k$ is a non-negative real number. The condition of the problem is equivalent to $xk = \frac{4x^{k}}{k} = 10$. If we multiply $xk$ by all the terms here, we get $(xk)^2 = 4x^{k+1}$. Since $xy = x^{k+1}$, $xy = \frac{(xk)^2}{4} = \frac{100}{4} = 25$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MathPerson12321
3749 posts
#42
Y by
It’s literally just log properties, 25 is the answer.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lpieleanu
3001 posts
#43
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Yusuf29
8 posts
#44
Y by
X^10=Y^x. Y^10=X^(100:x)=x^4y 4xy=100 xy=25
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cinnamon_e
703 posts
#45
Y by
solution
Z K Y
N Quick Reply
G
H
=
a