When I graphed this prior to playing the answer, I found that “a” was suspiciously close to pi, and was looking forward to that in the rest of the video. Turns out, it’s about 0.2% less than pi, so false alarm.
desmos says x=3.136 if anyone is curious. I threw it in Wolfram Alpha if you want more digits. x ≈ 3.13556453061975... Proof whether it's transcendental left as a homework exercise.
This can be done a lot quicker without doing the algebra: Note that from the start, sqrt(2) is an obvious solution. Furthermore, the LHS is a polynomial that is increasing for x>0 and the RHS is an exponential function with base >1. To see if there are further solutions plug in x=2>sqrt(2) which gives 4 on the RHS and the LHS is most definitely bigger than 4 so at 2 the polynomial is greater than the exponential. Since Exponential always beats polynomial there is exactly one more positive solution. We only have to figure whether that solution is bigger than 6-sqrt(2) or not. Note that when plugging in x=4, the RHS is 256 while the LHS is less than 4^4=196 so the second solution lies between 2 and 4. Thus the sum of the solutions is between sqrt(2) and 4+sqrt(2)
What does X equals 2>sqrt 2..u don't get that notation..that would mean 2 greater than srt 2..hiw is the mea jng of that clear? Did you mean to write x= 2^sqrt2? And why think to check then 6 minus sqrt 2...thanks for sharing..why not just check 6 or 4?
@@leif1075 I just wanted to point out that sqrt(2) is less than 2 because we needed to find out the begaviour of LHS and RHS after x=sqrt(2). So there is another solution bigger than sqrt(2). Thus the sum of positive real roots is also bigger than sqrt(2). Now we just have to find out if the sum is more than 6 or not and since we already know one root, that is equivalent to the second root being greater/smaller than 6-sqrt(2). And we know that 1
I didn't take logs, just worked with the given equations. For x >= 0 both are concave up, so they cross at most two times, and we know in the long term anything-to-the-x will dominate x-to-the-anything. We also know 2-to-the-root-2 is more than 2 and less than 4. Then by instection: : at x = 1, the left side = 1 while the right side = 2 (rhs dominates) : at x = root-2, the two are equal : at x = 2, the left side is between 4 and 16, while the right side is 4 (lhs dominates), so there's a second intersection : at x = 4, the left side is between 16 and 256 while the right side is 256 (rhs again dominates), so the second intersection is somewhere between x = 2 and x = 4 Knowing this, S is between root-2 + 2 and root-2 + 4, so D is the correct answer simply by inspection of the given equations.
Could you elaborate on how both being concave up means that they'll intersect only twice at most? I can't seem to prove anything like that, and seem to have found a counterexample that intersects at 3 points {-1, 0, 1}: y = x^4 and y = x^2
@@cigmorfil4101 y = x^2 + sinx and y = x^2 + cosx are both convex (concave up) and cross each other an infinite amount of times while not being always equal.
@@vladislav_sidorenko I was being a bit sarcastic with wanting a proof as it is only "obvious"[1] for the equations of the problem. y = (x-1)^4 and y = (x-1)^2 intersect 3 times when x >= 0... [1]obvious patterns need to be treated with care: take a circle, select a point on the circumference, it divides the circle onto 1 area. Select another point, draw the chord between the two, it divides the circle into 2 parts. Select another point, draw all the chords between it and the other points. The cycle is divided into 4 parts. Select a 4th point, draw all the chords between it and the other 3 points (the point is such that where any two chords meet *only* two chords meet), the area of the circle is divided into 8 parts Obviously the number of parts is 2^(points-1). So when a 5th point is added there will be 2^(5-1) = 16 parts? Nope it's only 15 (as a maximum)...
Saw the question, looked at the clock, and just figured out that at this time ill just watch ;) I know i should refresh my math memory with logs... but not at this hour ;)
I took log only once, spotted the obvious solution sqr(2), estimated 2^(sqr(2)+1) to be about 5.5, then plugged in some values to bound the remaining intersection point other than sqr(2) and got the same answer. At x=1 the LHS is greater than S must be greater than 2. On the other hand we have that 2^(sqr(2)+1)* log2(4)
No need for all this log manipulation ... sqrt(2) is an obvious solution. We also know that, for positive values, the left side starts at zero crosses with the right side at x=sqrt(2) but grows slower then the right side, so it will cross again. Therefore A and B are discarded. Then we can plot for x=6-sqrt(2) and see that the right side is already bigger than the left side, so the second solution is between sqrt(2) and 6-sqrt(2). Leaving only option D.
x^2^√2 = √2^2^x Define for non-negative values of x : f(x) = x^2^√2 ≤ x^2^(3/2) = x^(2√2) ≤ x^3 is a polynomial growth function. g(x) = √2^2^x = (2^½)^(2^x) = 2^(½*2^x)= 2^2^(x-1) is an exponential growth function. Obviously, one solution is x = √2 , because f(√2) = (√2)^2^(√2) = g(√2) . Now let's make a table of f(x) and g(x), for x = 0 ,1, 2, 3, 4, 5 x f(x) < x^3 g(x) = 2^2^(x-1) 0 0 √2 1 1 2 2
I did this similar to Presh, with similar reasoning, except I compared the two sides of the equivalent expression 2^(sqrt(2) + 1)log(x) = 2^x, x > 0 and the log is base 2. x < 1, LS < RS Left side negative, right side positive x = 1, LS < RS : 0 < 2 x = 2 , LS > RS : 2^(sqrt(2) + 1) > 4 x = 4 , LS < RS. : 2^(sqrt(2) + 2) < 16 X > 4, LS < RS, exponential grows faster than log expression I could have drawn a graph but this implies roots between 1 and 2, and then 2 and 4. So, with these bounds on the two values of x, the answer D is the only one that made sense. I didn't have to find any of the roots to this equation.
So, x = √2^2^√2 doesn't work? What becomes of the LHS and RHS if x is set to this? I thought that and √2^2^√2^2^√2 and all subsequent versions would all be solutions... edit: For the LHS it seems to go wrong. I think with (a^b)^c = a^(bc) and x = (a^b) = √2^2^√2, so a=√2 and b=2√2 and c=2√2 you get √2^(2√2*2√2) not the staircase of 5 powers. Is that where I went wrong?
One of the solutions x = √2 , in which case both lefthandside and righthandside become √2 ^(2 ^√2) ≈ 2.518512814 . x = √2 ^2 ^√2 does not work, because then the lefthandside becomes [√2 ^(2 ^√2)]^(2 ^√2) ≈ [2.518512814]^(2.6651441427) ≈ 6.7121996745 while the righthandside becomes √2 ^(2 ^[√2 ^(2^√2)]) ≈ √2 ^(2 ^[2.518512814]) ≈ √2 ^(5.7299113318) ≈ 7.2851347927 Note: a^b^c^d means a^(b^(c^d)) , and in general this does _not_ equal ((a^b)^c)^d , nor (a^b)^(c^d) .
No, again wrong: in your case in the lefthandside, b = 2^(√2) and c = 2^(√2) , which do _not_ equal 2√2 = 2^(3/2) . In your case, the lefthandside becomes (a^b)^c = (2^[2^(√2)])^[2^(√2)] = 2^( [2^(√2)] * [2^(√2)] ) = 2^( [2^(√2 + √2)] ) = 2^[2^(2√2)] (which can eventually be rewritten as 2^2^2^(3/2) , since 2√2 = √2 * √2 * √2 = (√2)^3 = [2^(1/2)]^3 = 2^(3/2) .) I hope that helps.
Someone please explain this. I used Microsoft math app to solve an equation, it showed the answer 9, then I clicked on edit but didn't change anything and returned for answer again. Now it showed the answer is 1. What's going on ? ua-cam.com/users/shortsKbrXtEDB5vI?si=4hY2YtJGj-tI8RlN
@@pandafanta How can i know it bro, because i am now only 16 yrs. I think this problem is 14-15yrs old teens. Its not that hard that everbody thinks. There is always a unique way, by which a problem can solved!!!!
@@chandranisahanone Well it appears i misunderstood your question, however i don't think the question would be for 14-15 year olds, i just feel like it is too difficult for that age
When I graphed this prior to playing the answer, I found that “a” was suspiciously close to pi, and was looking forward to that in the rest of the video. Turns out, it’s about 0.2% less than pi, so false alarm.
@@fhffhffwhat no
desmos says x=3.136 if anyone is curious. I threw it in Wolfram Alpha if you want more digits. x ≈ 3.13556453061975... Proof whether it's transcendental left as a homework exercise.
which 'a'?
This can be done a lot quicker without doing the algebra: Note that from the start, sqrt(2) is an obvious solution. Furthermore, the LHS is a polynomial that is increasing for x>0 and the RHS is an exponential function with base >1. To see if there are further solutions plug in x=2>sqrt(2) which gives 4 on the RHS and the LHS is most definitely bigger than 4 so at 2 the polynomial is greater than the exponential. Since Exponential always beats polynomial there is exactly one more positive solution. We only have to figure whether that solution is bigger than 6-sqrt(2) or not. Note that when plugging in x=4, the RHS is 256 while the LHS is less than 4^4=196 so the second solution lies between 2 and 4. Thus the sum of the solutions is between sqrt(2) and 4+sqrt(2)
What does X equals 2>sqrt 2..u don't get that notation..that would mean 2 greater than srt 2..hiw is the mea jng of that clear? Did you mean to write x= 2^sqrt2? And why think to check then 6 minus sqrt 2...thanks for sharing..why not just check 6 or 4?
@@leif1075 I just wanted to point out that sqrt(2) is less than 2 because we needed to find out the begaviour of LHS and RHS after x=sqrt(2). So there is another solution bigger than sqrt(2). Thus the sum of positive real roots is also bigger than sqrt(2). Now we just have to find out if the sum is more than 6 or not and since we already know one root, that is equivalent to the second root being greater/smaller than 6-sqrt(2). And we know that 1
These videos have helped me be better at maths, thanks
I didn't take logs, just worked with the given equations. For x >= 0 both are concave up, so they cross at most two times, and we know in the long term anything-to-the-x will dominate x-to-the-anything. We also know 2-to-the-root-2 is more than 2 and less than 4. Then by instection:
: at x = 1, the left side = 1 while the right side = 2 (rhs dominates)
: at x = root-2, the two are equal
: at x = 2, the left side is between 4 and 16, while the right side is 4 (lhs dominates), so there's a second intersection
: at x = 4, the left side is between 16 and 256 while the right side is 256 (rhs again dominates), so the second intersection is somewhere between x = 2 and x = 4
Knowing this, S is between root-2 + 2 and root-2 + 4, so D is the correct answer simply by inspection of the given equations.
Could you elaborate on how both being concave up means that they'll intersect only twice at most? I can't seem to prove anything like that, and seem to have found a counterexample that intersects at 3 points {-1, 0, 1}: y = x^4 and y = x^2
@@vladislav_sidorenko
But for x >= 0 your example curves intersect only twice.
However, it is an interesting claim of which I'd like to see a proof.
@@cigmorfil4101 y = x^2 + sinx and y = x^2 + cosx are both convex (concave up) and cross each other an infinite amount of times while not being always equal.
@@vladislav_sidorenko
I was being a bit sarcastic with wanting a proof as it is only "obvious"[1] for the equations of the problem.
y = (x-1)^4 and y = (x-1)^2 intersect 3 times when x >= 0...
[1]obvious patterns need to be treated with care: take a circle, select a point on the circumference, it divides the circle onto 1 area. Select another point, draw the chord between the two, it divides the circle into 2 parts. Select another point, draw all the chords between it and the other points. The cycle is divided into 4 parts. Select a 4th point, draw all the chords between it and the other 3 points (the point is such that where any two chords meet *only* two chords meet), the area of the circle is divided into 8 parts Obviously the number of parts is 2^(points-1). So when a 5th point is added there will be 2^(5-1) = 16 parts? Nope it's only 15 (as a maximum)...
It's actually 16. The pattern breaks with the next point only giving 31 separate areas.
6:47 Me at that exact moment notices the digits of rounded up *Pi* in order that is 3.1416 😂
Take logarithm by base square root of 2 and get S=2 on the base of convex considerations, see “Solve without pen”, problem 393 Repeated Exponents.
yeah, there's no way i would have gotten that!
But thanks to your very good explanation i was able to follow easily 👏
Wolfram Alpha groups the exponents different when you type in the equation as-is.
Great Question, great solution! Großartig! Vielen Dank
Saw the question, looked at the clock, and just figured out that at this time ill just watch ;) I know i should refresh my math memory with logs... but not at this hour ;)
I haven't tried solving it, but just by substituting x = √2, the equality holds good.
This video helps. I'm weak at inequality problems.
answer=(c) 2/
Nice graphical interpretation
I took log only once, spotted the obvious solution sqr(2), estimated 2^(sqr(2)+1) to be about 5.5, then plugged in some values to bound the remaining intersection point other than sqr(2) and got the same answer. At x=1 the LHS is greater than S must be greater than 2. On the other hand we have that 2^(sqr(2)+1)* log2(4)
No need for all this log manipulation ...
sqrt(2) is an obvious solution.
We also know that, for positive values, the left side starts at zero crosses with the right side at x=sqrt(2) but grows slower then the right side, so it will cross again. Therefore A and B are discarded.
Then we can plot for x=6-sqrt(2) and see that the right side is already bigger than the left side, so the second solution is between sqrt(2) and 6-sqrt(2). Leaving only option D.
How is S = root(2) + a 5:58
Direct answer compire it X is equal to square root of 2
Very nice one, reminded me of power tower problem also involving square root of two.
How can you solve something like this under time pressure if you haven’t seen the solution?
What a tough question, worth of an Olympiad Team selection test.
nope even just USAMO qualifiers should be able to do this in under 5 minutes
Take log with base2 and arrange as this.
f(x)=k, k is 2 oo
x = 2 , f(x) = 2
x = 4 , f(x) = 4
so, 1< x1
Lambert W function could have found the second solution easily, it’s easy to solve this using Lambert W function.
Interesting method. I just used pure calculus without a calculator.
x^2^√2 = √2^2^x
Define for non-negative values of x :
f(x) = x^2^√2 ≤ x^2^(3/2) = x^(2√2) ≤ x^3 is a polynomial growth function.
g(x) = √2^2^x = (2^½)^(2^x) = 2^(½*2^x)= 2^2^(x-1) is an exponential growth function.
Obviously, one solution is x = √2 , because f(√2) = (√2)^2^(√2) = g(√2) .
Now let's make a table of f(x) and g(x), for x = 0 ,1, 2, 3, 4, 5
x f(x) < x^3 g(x) = 2^2^(x-1)
0 0 √2
1 1 2
2
Log2(log2(x)) has an asymptotic in x=1, not in x=0. Also it is not defined for x
i think we could make a function for the multiplier in relation to the exponent
I followed for about 7 seconds then my head exploded.
The graph not being exactly to scale bothers me probably alot more than it should
Very cool
Can any one solve the exact value of second solution??????
I refuse to accept the second solution, I need an accurate, like the first! 😁
I don't think the second solution has a algebraic form.
I did this similar to Presh, with similar reasoning, except I compared the two sides of the equivalent expression 2^(sqrt(2) + 1)log(x) = 2^x, x > 0 and the log is base 2.
x < 1, LS < RS Left side negative, right side positive
x = 1, LS < RS : 0 < 2
x = 2 , LS > RS : 2^(sqrt(2) + 1) > 4
x = 4 , LS < RS. : 2^(sqrt(2) + 2) < 16
X > 4, LS < RS, exponential grows faster than log expression
I could have drawn a graph but this implies roots between 1 and 2, and then 2 and 4. So, with these bounds on the two values of x, the answer D is the only one that made sense. I didn't have to find any of the roots to this equation.
Vow.. thats the shortest 8m video... felt like 1min
So, x = √2^2^√2 doesn't work?
What becomes of the LHS and RHS if x is set to this?
I thought that and √2^2^√2^2^√2 and all subsequent versions would all be solutions...
edit:
For the LHS it seems to go wrong.
I think with (a^b)^c = a^(bc) and x = (a^b) = √2^2^√2, so a=√2 and b=2√2 and c=2√2 you get √2^(2√2*2√2) not the staircase of 5 powers.
Is that where I went wrong?
One of the solutions x = √2 , in which case both lefthandside and righthandside become
√2 ^(2 ^√2) ≈ 2.518512814 .
x = √2 ^2 ^√2 does not work, because then the lefthandside becomes
[√2 ^(2 ^√2)]^(2 ^√2) ≈ [2.518512814]^(2.6651441427) ≈ 6.7121996745
while the righthandside becomes
√2 ^(2 ^[√2 ^(2^√2)]) ≈ √2 ^(2 ^[2.518512814]) ≈ √2 ^(5.7299113318) ≈ 7.2851347927
Note: a^b^c^d means a^(b^(c^d)) , and in general this does _not_ equal ((a^b)^c)^d , nor (a^b)^(c^d) .
No, again wrong: in your case in the lefthandside, b = 2^(√2) and c = 2^(√2) , which do _not_ equal 2√2 = 2^(3/2) .
In your case, the lefthandside becomes
(a^b)^c = (2^[2^(√2)])^[2^(√2)] = 2^( [2^(√2)] * [2^(√2)] ) = 2^( [2^(√2 + √2)] ) = 2^[2^(2√2)]
(which can eventually be rewritten as 2^2^2^(3/2) , since 2√2 = √2 * √2 * √2 = (√2)^3 = [2^(1/2)]^3 = 2^(3/2) .)
I hope that helps.
RAVI SIR🎉🎉🎉
So this 8 minute video could be done in 15s. Just use a graphing utility.
x 2/2=/2 2x
Someone please explain this. I used Microsoft math app to solve an equation, it showed the answer 9, then I clicked on edit but didn't change anything and returned for answer again. Now it showed the answer is 1. What's going on ? ua-cam.com/users/shortsKbrXtEDB5vI?si=4hY2YtJGj-tI8RlN
You can use Geogebra and find the 2 solutions : sqrt(2) and around 3,1355.
Good One.
You lost me 5/6 of the way thorugh.
Don't read comments 😂
Why you read it "x to the power of 2?" It should be read "x square raise to the square root of 2!"
No, it shouldn't.
Replace x with sqrt(2) and the LHS = the RHS.
Option E is also true, guess you wanted x ≥ 6 but you wrote x ≤ 6.
Please carefully look at the options. He has written 6 less than equal to S.
Math is not annoying
MATH IS TERRIBLE
Plzz presh talwalker reply me once i am your big fan and supporter since 2021❤❤❤
Pls teel this question is for which class
@@pandafanta It is math olympiad qulifier question lmao not for regular class
@@pandafanta How can i know it bro, because i am now only 16 yrs. I think this problem is 14-15yrs old teens. Its not that hard that everbody thinks. There is always a unique way, by which a problem can solved!!!!
@@kohlsnofl5110Yeah it's true but for what level of age's it is based it's not mention. So i just make a guess.
@@chandranisahanone Well it appears i misunderstood your question, however i don't think the question would be for 14-15 year olds, i just feel like it is too difficult for that age
But you were asked to answer for the SUM of the solutions.
And the answer in the video does indeed identify the range of the SUM of solutions. Watch to the end!
So, the solution is around sqrt (2)+3,1355 !!