Solving a Nice Exponential Equation from Romania

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#ChallengingMathProblems #ExponentialEquations

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[stickmanbattle]

divide both sides by 2^([2x-1]/[x-1]) you can easily solve the problem.

[Icewallocumm]

Nice!


Aceasa ecuatie exponentiala este chiar interesanta.

[noeticresearch]

Very interesting math problem, taken from what looked like complexity to simplicity SyberMath. Thanks very much! 👍 👌

[olivierklepper]

Your graphic writing is getting ever better! Nice work!

[armacham]

substitute y = x - 1, that makes the problem a lot simpler
simplify the fractions, and then substitute z = 1/y
and again, the problem becomes even easier

[rrivierareject]

An alternative I tried: those rational exponents are ultimately going to be integer quotients with rational remainders (e.g., (2x-1)/(x-1) = 2+1/(x-1), ...)

That cleaned things up very nicely--one bit of factoring and a simple enough proportion was all I needed.

[marcushendriksen]

The only pair of powers of 2 that equal 24 is 8 and 16. That means either 2x-1/x-1 or 3x-1/x-1 is 4 and the other is 3. Since they have the same denominators, the one with the bigger numerator will be the bigger one overall, so 3x-1/x-1 must be 4. That means x must be 2. Bada bing bada boom

[johablima]

I did in a slightly harder way.
First I divided the polynomials, getting 2^(2+1/[x-1]) + 2^(3+1/[x-1]) = 24, then i separated the exponents:
2² + 2^(1/[x-1]) + 2³ + 2^(1/[x-1]) = 24
dividing everything by 2², we have:
2^(1/[x-1]) + 2×2^(1/[x-1]) = 6
3×2^(1/[x-1]) = 6
2^(1/[x-1]) = 2¹
1/(x-1) = 1
x-1 = 1
x = 2

[stevenlitvintchouk]

I assumed only real roots. Just by inspection, you can see that 2^3 + 2^4 = 24 will work, meaning that x = 2 is a root. The first exponent can be rewritten as 2 + 1/(x-1), and the second exponent can be rewritten as 3 + 1/(x-1). So for X>2, as X grows to infinity, the exponents approach 2 and 3 respectively, so the value of the expression declines from 24 to 12, so there are no more roots there. Likewise for X < 1, as X goes to negative infinity, the value of the expression also goes to 12. For X = 1, the expression is undefined; no root there. For 1 < X < 2, as X approaches 1, the exponents both grow from 3 and 4 respectively to infinity, so the expression goes from 24 to infinity, so there can't be any more roots there either.

[tsf]

16 and 8 are only powers of 2 which sum up to 24 hence the powers can be equated to 4 or 3

[КатяРыбакова-шд]

Если из каждой дроби выделить целую часть, получается простенькое уравнение с корнем 2. Ответ:2.

[Murlica]

It's easier to solve by representing 24 as 2^4+2^3 and notice that equalling the powers of the terms on both sides. It gives x=2 in the end

[EdgarVerdi]

I just changed y = x-1 and the problem quickly gave the answer by itself (factoring 2^(1/y) out of each term)

[MisterPenguin]

I've leveled up a lot, but my largest knowledge gap seems to be identifying things that are similar.

[avdhutpawar]

Dude just replace (2x-1) /(x-1) by t it will be very easy to solve. Equation becomes
2^t + 2^(t+1) = 24
On Simpliyfing equation becomes
2^t + (2^t . 2) = 24
Taking 2^t comman we get
2^t(1+2) = 24
2^t(3) = 24
2^t = 8 =2^3
On comparing both sides we will get t=3. Then equate (2x-1) /(x-1) = t = 3
On rearranging this equations you will get
2x-1 = 3x-3
Therefore you will get x=2.

[Rbmukthegreat]

Guessed that it would end up looking like 8 + 16, then noticed that 3x-2 is probably going to be bigger, so set (3x-2)/(x-1)=4 and got the answer from there! Now time to watch and see how it was meant to be done…

[chikodiigiri]

Please, how did you manipulate the exponents ?

[javadhamzavi]

2x-1/×-1 =2 +1/×-1 and 3x-2/×-1 = 3 +1/x-1 then you can factor out 2**1/x-1 that's easy and fast.

[kaarthikananthanarayanan]

Lots of love from India. I'm simply addicted to your YT channel.

[itsjustme]

Let first exponent be a and second be b, WLOG.
2^a + 2^b = 8 + 16
2^a + 2^b = 2^3 + 2^4

(a=3 and b=4) OR (a=4 and b=3)

a=3 and b=4 will lead to the solution for x