Two Ways to Solve | Bulgarian Mathematics Olympiad

[moeberry]

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.

[andreben]

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<0, set y=-x and the equation becomes
3^y * (2^y +1)=2^(2y+1)
which doesn't have a solution, since the LHS is odd and the RHS is even (just like in Adrian D's comment).

[ihatealgebra]

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!

[Horinius]

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

[blankmedia]

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

[abraham]

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

[mrmathcambodia]

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

[mathcanbeeasy]

It works identically with odd and even for the case X<0. When you substitute y=-x, for the case when X<0, you can do some calculations and reduce to 3^y(2^y+1)=2^(2y+1). So odd*odd=odd in LS and in RS we have even. So no solution for X<0.

[mcwulf]

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.

[basil-vander-elst]

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

[dhriti]

(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

[mcwulf]

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

[andreweberlein]

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).

[christopherellis]

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

[benmartinez]

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

[yt-]

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

[anurag]

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

[spencergee]

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

[satrajitghosh]

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

[AEROXYDHYA]

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