BULGARIAN MATHEMATICAL OLYMPIAD | Algebra Problem

BULGARIAN MATHEMATICAL OLYMPIAD | Algebra Problem

Find the value of x
(8^x - 2^x)/(6^x - 3^x) = 2

[stevenwilson]

I got the x = 1 trivial solution by inspection, the magic of this video was proving this was the only solution using parity. Well done.

[tzortzos_aristotle]

I read everyone in the comments (including myself when I watched the thumbnail) saying how fast and easy they got the answer. The whole point of the video is not just to prove that the solution is x=1, but also to make sure it’s the only solution

[martynasbendikas]

I have no idea how this video was suggested to me... but damn i needed this! I used to be so great back at school at solving these... but 12 years passed and i keep thinking about how i wouldn't even know where to start if i were given a simple math question to solve. This one reminded me a lot. Not just formulas, but also the excitement of solving it. Thank you! Will definitely check more videos!

[hanswust]

Wonderful solution!

Unfortunately, some people don't understand that you have to follow a rigorous method to call your findings a solution for trial and error guessings are not a mathematical procedure.

[icydingo]

Glad to see that problems from Olympiads from my country are reaching this far!

[eon]

Every time I see an equation like this, I’m like “Let’s obviously start with 1”

[quantgeekery]

The limit to integers makes the x=1 a solution that jumps out. The proof of no other solutions is the beauty of logarithms.

[johnwiese]

I started with 2, which was 60/27, which is above 2, then tried 1 to see if it got closer or farther than 2. 1 ended up being correct.

[yairkaz]

Interesting thing is if we graph both equations we see that they have two points in which they collide, one of them is 1 which fits the equation and the other one is 0 which does not fit the equation as 0/0 is undefined

[toshangupta]

Alternate method :-
8^x-2^x/ 6^x-3^x =2
2^(3x)- 2^(x) / 2^x 3^x -3^x= 2
2^x(2^x+1)/ 3^x =2 (x≠0)
(2/3)^x (2^x+1) =2

Now if we check whether LHS is increasing or decreasing function then
Let f(x)= (2/3)^x (2^x+1)
f'(x)= (2/3)^x ln(2/3) (2^x+1) +(2/3)^x .2^x ln 2
= (2/3)^x[ (2^x+1) ln (2/3) + 2^x ln2 ]

Now a^x is always ≥0 for all X
So f'(x)≥0 for all x belonging to its domain .
So f(x) is increasing function in its domain

Once f(x)=2 it would never be equal to 2 again in its domain
By observation we can see at x=1 it is equal to 2

So x=1 is the only solution

[thomaswagner]

parity is something i never would’ve thought to use. i was thinking trying to use log rules and e^ln. very cool!

[abategameryt]

I only ingnored the x to see if x was 1, so i resolved it in ~10 seconds xD

[mrvnoble]

Cool use of parity. The only part I got confused by was using I as the domain instead of Z! 😋

[Mr.Exquisite]

Plot it and you will find 0 is another solution. Just by performing a x≠0 division step does not rule out the possibility, just makes it impossible to prove when going further.

[NishantCosmos]

i got x =1, took me 20secs
using my secondary level math skills
and this guy wrote whole syllabus to solve this equation.

[tawhv]

For 2^x(2^x+1) / 3^x = 2
x = 0 is a second solution.

(8^x-2^x)/(6^x-3^x) is undefined because 0/0. The limes for x → 0 is 2.

[gholmes]

I used the Bowdler reduction applied to Von Gaggentrauzers theory of longness but still ended up breaking my drill bit.

[l.w.paradis]

I love stuff that proves a solution is unique. Cool stuff.

[MironParshikov]

After 2^x(2^x+1) = 2*3^x, we can solve 2 two systems:
1) {2^x = 2; 2^x+1 = 3^x} from which we get x = 1 using log() with base = 2
2) {2^x = 3^x; 2^x + 1 = 2} => 2^x = 1 => using log() with base = 2 we get x = 0 that doesn’t satisfy our first restriction
=> x = 1

[gibraltar]

I saw it and said “yeah it’s 1” and I was right! I’m glad my college math is serving me well