Two Ways to Solve | Bulgarian Mathematics Olympiad

[@moeberry8226]

Method 2 is a much stronger statement it shows that not only there is no solution after or before x equals one with respect to the integers but it shows it’s true through out all real numbers.

[@cobokobo2115]

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[@toppinzr3743]

Actually true for all x complex numbers. Once you factor out 2^x - 1, you have the equation 4^x + 2^x = 2 3^x, or 4^x - 3^x = 3^x - 2^x. i.e. the value of (y+1)^x - y^x is the same at y = 2 and y = 3. By the mean value theorem, this means that x(y+1)^(x-1) - xy^(x-1) = 0 for some y in (2,3). Since x=0 can't be a solution except as a limit, you can factor out x. You get (1+1/y)^(x-1) = 1, for some y in (2,3). This can only happen if x-1 = 0. So x=1.

[@andreben6224]

Hmm 🤔 when you are at the level of 2^x * (2^x +1)=2 * 3^x you could simply use the unique factorization theorem of integers and settle that 2^x=2 2^x +1 = 3^x thus x=1. When x

[@cobokobo2115]

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[@curtisclement5467]

Nice!

[@Horinius]

The first method is way more elegant than the 2nd one. Nice job 👍

[@ihatealgebra2431]

This is my solution: 8^x - 2^x | 6^x - 3^x = 2 Rearrange to get (4^x - 1)2^x | (2^x - 1)3^x = 2 and simplify further so the numerator on the left side is (2^2x - 1)2^x Multiply both sides by 3^x | 2^x to get 3^x | 2^x-1 on the right side Now, we just create two new equations as both sides are fractions. Rearrange both equations to equate zero and then equate the two equations together to get 2^x - 1 - 2^x-1 = 2^2x - 1 - 3^x Isolate 3^x 3^x = 2^x-1 + 2^2x - 2^x In order for any values of x to satisfy the above equation, the right expression has to be A PERFECT CUBE. We know that adding, multiplying, and subtracting even numbers will always equate even numbers. However, except for such case where one term is to the power of zero which would equate an odd number: 1. In order for such case to happen, x can be equal to 1 and 0. By plugging in both numbers, the only expression that equates A PERFECT CUBE is for x to be equal to 1. Boom! Solved!

[@mathcanbeeasy]

It works identically with odd and even for the case X

[@cobokobo2115]

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[@blankmedia01]

I hate being addicted to this 😒 bahasa it's Saturday and here I am consuming math

[@abraham4124]

The question was deceptively simple. Thanks for sharing the solution 👍

[@cobokobo2115]

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[@mcwulf25]

Came back to this, forgetting I had previously dont it! Pretty much how you did it. Except that for the final stages of the first solution I kept the factors separate, ie 2^(x-1) * (2^x - 1) = 3^x requires that both factors on the LHS must be odd. So x = 1.

[@andreweberlein1509]

Let f(x) be the LHS defined on the real numbers. Observation 1: x=0 is a removable discontinuity, so I choose to include f(0)=lim f(x) as x->0. Observation 2: f(x) is everywhere concave up. Observation 3: f(0)=f(1)=2 Therefore, x=0 and x=1 are the only solutions (probably by intermediate value theorem in some capacity).

[@mcwulf25]

Haven't looked at the video yet but I solved it by: - factoring out (2/3)^x - factoring numerator as difference of squares - cancelling 2^x - 1 - cancelling 2 each side - multiply by 3^x, RHS is odd. - LHS is even if X>0 so x=0

[@cobokobo2115]

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[@mcwulf25]

I mean x>1 so x = 1

[@dhriti2209]

(8ˣ-2ˣ)/(6ˣ-3ˣ)=2 2ˣ(4ˣ-1)/3ˣ(2ˣ-1)=2 2ˣ(2ˣ+1)/3ˣ=2 2ˣ(2ˣ+1)=2(3ˣ) Obviously 2ˣ+1 is odd and thus has to be equal to 3ˣ (also odd). Thus 2ˣ=2 and 3ˣ=2ˣ+1 x=1 only solution

[@mrmathcambodia2451]

so good , I like this exam , keep sharing more .

[@yt-1161]

I got (4/3)^x + (2/3)^x =2 then divided this into cases, used Bernoulli inequality and kind of ruled out negative case (invert the fractions) with the following argument: if you denote 1st term (1-y) and 2nd term (1+y) then their product ist always less than 1 but (9/8)^x ist always greater than 1. I had solved this long time ago

[@HiHi-gy6vy]

6;06 A different approach:- 2^x(2^x+1)=2×3^x So 2^x=2 Hence 2^x=2^0 x=0 , 2^x+1=3^x Multiplying by log log(2^x+1)=log3^x log2^x × log1 = 3logx Since log1=0 0=3logx 0=logx X=1 Therefore x=0,1

[@nirmalkumarsingh1829]

The use of log was amazing

[@two697]

That's not how logarithms work

[@cobokobo2115]

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[@cobokobo2115]

@@nirmalkumarsingh1829 amazing .......kzbin.info/www/bejne/rqCWeIFmd6iojtk

[@satrajitghosh8162]

A bit of work shows if 2^x -1 0 then (4^x + 2^x)/(3^x) = 2 or 4^x + 2^x = 2* 3^x The trivial solution, x= 0 contradicts to 2^x -1 0, so unacceptable A feasible solution is there at x=1. At x > = 2, LHS grows faster than RHS

[@xd_metrix]

I'am stupid, but why is 2^x * 3^x = 6^x, please.

[@Gringohuevon]

nobody does maths like this

[@78anurag]

Wow, we had the same solution! (The first one)

[@spencergee6948]

Maths is maths. It doesn't matter where it originates. It would be exactly the same problem had it come from Barnsley.

[@atpugnes]

The second derivative test seems to be redundant

[@cobokobo2115]

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[@AEROXYDHYA]

I have much nice idea which solves this problem in 6 second

[@basil-vander-elst]

Just tried x= 1 and turned out to be right hehe

[@benmartinez1267]

I looked at it and thought x=0 and saw can’t be. Then thought x=1 and it worked. Done in 3 seconds.

[@mojota6938]

The hard part is to show there are no other solutions.

[@cobokobo2115]

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[@cobokobo2115]

@@mojota6938 amazing .........kzbin.info/www/bejne/rqCWeIFmd6iojtk

[@Thorleif13]

I just expected x to be one and it worked

[@VIKASVERMA-nk8uw]

Great

[@cobokobo2115]

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[@thedeadbaby]

guessed 1 is a solution. QED? j/k i know they don't take correct guesses and answers.

[@broytingaravsol]

this answer can be observed by the formula

[@mcwulf25]

But can you prove it's the only answer?

[@cobokobo2115]

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[@beybld]

@@mcwulf25 one answer is good enouf to me

[@murthys743]

could you please improve the audio quality

[@cobokobo2115]

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[@venum8259]

Put x=1 We get 8^1-2^1 ÷ 6^1- 3^1 = 6÷ 2 = 2 When x=1 this becomes true. Hence x=1 is the answer. This is trial and error method.

[@supercool1312]

6/2 is not 2 what

[@venum8259]

@@supercool1312 printing error It should be 6÷3 which is equal to 2. Immediately after I printed it I realised the mistake of my finger. It is not the mistake of knowledge. You see the method only. The method is very simple and easy to understand.

[@christopherellis2663]

6-3=3 8-2=6 6/3=2 ×=1

[@christopherellis2663]

It is simply that x=1. Because 6/3=3 I don't see the point of the elaborate dissection.

[@andrea-mj9ce]

The point is to show that there's no other solution

[@geneautry2091]

What's wrong with me? I had that answere in less than 10 seconds. I understand you're trying to make xtra cash stretching explanations out.

[@advaykumar9726]

Copied from Flammable Maths

[@tianqilong8366]

dont know if it is a copy paste, but flammable maths does post this same question 2 days ago hahaha

[@rahmatmatematika6534]

Oh..., Bulgarian IMO.

[@jfcrow1]

Flammable Maths is a joke. He even said so. No Genius IQ. All lies. just regurgitates solutions.

[@cobokobo2115]

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