Math Olympiad | A Nice Exponential Problem | 90% Failed to solve !

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[@daniellopez3124]

Muy buen ejercicio, lo he disfrutado muchísimo. Gracias profesor, un saludo desde Lima, Perú

[@neerarajpoot8409]

It can be solved by putting x=0,1 or 2. Then solution will be x=2. The short method.

[@vijaymaths5483]

Thank you for your feedback!! 😊

[@LeakyFaucett]

At minute 6:39 you have 6 sqrt2 divided by 2. It should be 3sqrt2 for the second half of the sum, not 2sqrt2. Hope this helps.

[@prabhushettysangame6601]

4 square root 2 divided by 2 means 2 square 2 ( square root 32 = 4 square root 2 )
Please recheck once again that part 🎉

[@vijaymaths5483]

Ok hi hai, aur ek baar check kijiye

[@WhiteGandalfs]

(7:35) What you do at this point is an auguration of the solution. This has only a use in one (of endlessly many) cases - if the value of x IS IN FACT 2.
BUT: IF you DO augur the end result (in that you manipulate expressions in a way that has no allowance except the one and only case where the augurated result does apply), then you could have simplified the whole matter drastically, in that you could have tested the augurated result in the original expression.
In fact, first i have solved this case in that way - trying to imagine an approximate value (which is trivial at first glance "about 2"), then testing how far away from the destination that would be - only to see that it is actually a solution. And since the expressions sqrt(2)+1 and sqrt(2)-1 are invers to each other, it was as well trivial to deduce that a second solution must exist with negated value.
To solve that problem in a generalized way, you have to take logarithms.
Let's say the LHS should equal an arbitrary K.
Then you get the t in your midway approach as t = k/2 +- sqrt( k^2 / 4 - 1 )
Then the generalized solution is log( t ) / log( sqrt(2)-1 )
...for t1 and t2.

[@vijaymaths5483]

Got it... Thank you for your feedback master👍👏

[@shmuelzehavi4940]

But all the point with this problem is of being a special case and not a general one. Moreover, you don’t have to augur the end result from the start, but just to find out that the intermediate result (3±2√2) equals to (√2+1) raised to an integer power (±2) .

[@johnnyfarfan1723]

Excelente trabajo. Un abrazo. Saludos.

[@vijaymaths5483]

Thank you ⚘️

[@ezzatabdo5027]

Thanks too much sir, I'm following yourself by that excellent nice and beautiful problems, thanks again.

[@vijaymaths5483]

Thanks a lot 🎈

[@antonio9633]

Excelente didática professor.

[@vijaymaths5483]

Thank you 🙏😊

[@juliovasquezdiaz2432]

Gracias por compartir el vídeo y la solución del ejercicio. Saludos desde Chiclayo Norte del Perú.

[@vijaymaths5483]

Thanks too for your valuable feedback 🖍

[@jaju2178]

Guess one solution x=2. a^x is strictly monotonically increasing for a>0, so x=2 is the only real solution. Finish.

[@ncslovers3429]

Nicely explained!

[@vijaymaths5483]

Thanks!

[@yasharhashimov9063]

Thank you for clear explaining

[@vijaymaths5483]

Welcome 😊

[@user-nd7th3hy4l]

X=2

[@Danieswors]

SENSACIONAL!!

[@vijaymaths5483]

Thank you 🙏

[@user-ri6rn7ti5h]

(x+2x-3)

[@superiorlyrics8326]

👏👏👏

[@vijaymaths5483]

🙏🙏

[@user-ee7nw2rx9s]

Этот пример уже был решен другим автором.

[@adroaldoferreira978]

Great.

[@vijaymaths5483]

Lot of thanks

[@sumit-mn6ys]

Great explanation 👍

[@vijaymaths5483]

Glad you liked it

[@user-ri6rn7ti5h]

(x+3x-3)

[@vijaymaths5483]

🤨

[@wRAAh]

Squiroot

[@comdo777]

asnwer=3 isit

[@vijaymaths5483]

No! Answer is 2 & -2...Thanks for watching 😊

[@jaggisaram4914]

2

[@vijaymaths5483]

Correct.. thanks for watching 🙂

[@klopklop6055]

молодец, садись, 4ре)

[@AvinashKumar-yz2tf]

X=2