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9^x+9^x=54 2(9^x)=54 9^x=27 9^x=3^3 (3^2)^x=3^3 X(log3^2)=log(3^3) 2xlog3=3log3 2x=(3log3)/log3 2x=3 X=2/3

-3😂

M=-3

m=-3

simple and easy solution one step only

M

If you are just considering positive integers, then why is x = 11, y = 0 a solution? 0 is not a positive integer.

This equation has infinitely many solutions.... All you did was get integer solutions by guessing and plugging into a set of linear equations but this is by no means a complete set of solutions... x=+-sqrt(121+y^2) such that y can be any freely chosen real number is the full form of the solution on the set of real numbers.

Very helpful

It was helpful thanks so much

I take my hat off to you my brother! This absolutely amazing! God bless you!

An excellent explanation. Thank you!

M= -3

What is the need for such functions? Will the society benefit from such functions? Does it represent some non linearly expanding Universe?!?

only m = -3. m =3 and m = 4 are extraneous

m = -3, 3, 4

"quickly and effectively" you say. Rubbish I say. Even ignoring the infinite number of solutions where the variables are not integer such as x=0, y=0, z=log to base 2 of 54. There are 6 different integer solutions, so your long-winded method 'proving' x=3 is more probably wrong as there are twice as many alternative answers, namely x=4 or x=5. x, y and z are indeed 3, 4 and 5, but the order cannot be determined with the information provided. You fail.

No complicated math is needed to solve the problem if the solution are entirely integers. As well x, y and z has to be 5 or less (as 2^6 = 64) Let’s try setting z to 5: 2^x + 2^y + 2^5 = 56 This means 2^x + 2^y + 32 = 56 leading to 2^x +2^y = 24. Both x and y must be less than 5 (as 2^5 = 32). Let’s try setting y to 4: 2^x + 2^4 = 24. This means 2^x + 16 = 24 leading to. 2^x = 8. It is now easy to see that x must be 3 (as 2^3 = 8). The three exponents (x,y,z) are 3, 4 and 5. It is not possible to say “which is which”, any combination will work. The method might seem complicated, but is done by head is seconds. 😁

3

56 in Boolean 111000. Ans 5,4 3.

Finally everything is correct but 34+32=66 and 56 is correct

It took me like 10 seconds to figure out the 32 + 16 + 8 = 56 so we have 3,4,5 I guess the main question is, is there some analytic way to figure out problems like this when the numbers are much larger... Since two to the power of 6 is 64 it is a fairly straightforward matter to check and guess numbers under six. Try setting up the same type of problem but replace 56 with a 12-digit number

-3 ?

56=32+16+8 =2^×+2^y+2^z =>(×=5>y=4>z=3) An assimulation only

Just work out the equation.I hope you dont teach young students because you are the type of teachers who make working out mathematical problems a nightmare. Talking too much explaining nothing.

Assuming that x is a positive integer, it takes about 10s to realize that a solution is (x, y, z) = (3, 4, 5).

He said 675 but wrote 625.😀

The answer is -3.

I feel like 2x speed and muted would make this more clear

The US olympiads seem easier than ones from other countries. I did this in my head and I'm not that good. I think it's easier if you write 81^m as 3^4m at the start, then repeatedly square both sides until you get 3^15 = 3^64m so 15 = 64m so m =15/64

From the initial inspection, it is clear that the unknown number m, is 3.

-3

m - 2x + N

Hmm yes... Harvard wannabes to demonstrate solving a cubic with a quadratic and already comfortable working with complex numbers ! 20 mins fair enough for a young student. What will be left for Harvard teach them ? Those that only focus on M1 need to brush up on the invention of imaginary numbers ! Thankyou...step 1 so critical !!!

Simple to solve according to law

m^2(1-m)=36 1-m=4(as 9×4=36) m=-3

Answer is M = -3. Took me less than 90 seconds using my brain.

In fact, this is systemically correct!!

There is a much faster solution: You see that m must be -3 (because (-3)^2 - (-3)^3 = 9 - (-27) = 36)! Maybe there are other solutions as well because this is a cubic equation.

This is complication not soloution

Terrible explanation. This is easy.

m = -3

After all that, you didn't even check that the complex solutions were also correct?

The answer is "-3". -3 x -3 = 9 -3 x -3 x -3 = -27 9 - (-27) 9 + 27 = 36

m³-m²+36=0 (m+3)(m²-4m+12)=0 m = -3 m²-4m+4+8=0 (m-2)²=-8 m = 2√2i+2 , -2√2i+2

Ya, like Fermat......

Great thanks for a nice, clear explanation. Loved it.

Once you find that m=-3 is one of the solutions, you can put the equation as (1): m^3-m^2+36=(m+3)(m^2+am+12)=0,a∈R ---(1) <=> m^3-m^2+36=m^3+(a+3)m^2+(12+3a)m+36=0,a∈R <=> -1=a+3,0=12+3a,a∈R <=> a=-4 ---(2) (1)(2) <=> m^3-m^2+36=(m+3)(m^2-4m+12)=0 This method is useful when you are having trouble factoring.

You say it is a Harvard Entrance Exam, but do I really need to write so much? Because I found all three roots just in 30 seconds in my head. Can I just write the answer?

5:18 I've lost the will to live ... Byeeee.