I just set the exponents equal to each other, which gives a cubic in x. The obvious solution is x = 1, but then the obvious solution to the original problem is also x = 1. Sometimes laziness works, especially when you're not as smart as a Romanian 10th grade student!
Great problem! Another way to show that x=1 is the only solution is to look at the first and second derivatives of f(x) = 2^(x-1) + 2^(1/sqrt(x)). First, note that any real solution of f(x) = 3 requires x>0, because negative x causes f(x) to become complex. f'(x) = log2*(2^(x-1) - 1/2*x^(-3/2)*2^(1/sqrt(x))), so f'(1) = log2*(2^0 - 1/2*1*2^(1)) = log2*(1 - 1) = 0, and f"(x) = (log2)^2*2^(x-1) + 3*(log2)*x^(-5/2)*2^(1/sqrt(x))-2) + (log2)^2*x^3*2^(1/sqrt(x)-2). We know that f"(x) must be positive because each of its terms is positive for all x>0. Therefore, f(x) is minimized at x=1, the only root of f(x) = 3.
Excellent approach by you sir never thought of it . i got to an expression like 2^x [1+2^(x+rootx-2x^2)] = 6 = 2 * 3 , from there i took 2^x = 2 and got x = 1 , i tried solvingx - rootx + 1 = 0 but idt im getting an answer . Can u confirm if my approach is somewhat valid ( i assumed integer answer to make that above assumption)
What does solve mean? My solution after 10 seconds is 3 = 2^0 + 2^1 means that x = 1. (It has been 65 years since I was in 10th grade, so my math may be a bit fuzzy.)
Can we use the straightly increasing or decreasing function before or after a certain values to solve this equation. It is obvious that 1 is a solution. We can test the function around this value maybe. 😉😉😉😉😉😉
I just set the exponents equal to each other, which gives a cubic in x. The obvious solution is x = 1, but then the obvious solution to the original problem is also x = 1. Sometimes laziness works, especially when you're not as smart as a Romanian 10th grade student!
Great problem! Another way to show that x=1 is the only solution is to look at the first and second derivatives of f(x) = 2^(x-1) + 2^(1/sqrt(x)). First, note that any real solution of f(x) = 3 requires x>0, because negative x causes f(x) to become complex. f'(x) = log2*(2^(x-1) - 1/2*x^(-3/2)*2^(1/sqrt(x))), so f'(1) = log2*(2^0 - 1/2*1*2^(1)) = log2*(1 - 1) = 0, and f"(x) = (log2)^2*2^(x-1) + 3*(log2)*x^(-5/2)*2^(1/sqrt(x))-2) + (log2)^2*x^3*2^(1/sqrt(x)-2). We know that f"(x) must be positive because each of its terms is positive for all x>0. Therefore, f(x) is minimized at x=1, the only root of f(x) = 3.
Actually, f(x
Beautiful
Thank you
This was the first question on this year's district round 10'th grade math Olympiad.
Also, nice to mention Romania on our national day.
La mulți ani!
Mă bucur să aud asta. Mulțumesc!
I also got x=1 as the only real solution.
Excellent approach by you sir never thought of it . i got to an expression like 2^x [1+2^(x+rootx-2x^2)] = 6 = 2 * 3 , from there i took 2^x = 2 and got x = 1 , i tried solvingx - rootx + 1 = 0 but idt im getting an answer . Can u confirm if my approach is somewhat valid ( i assumed integer answer to make that above assumption)
Excellent
What does solve mean? My solution after 10 seconds is 3 = 2^0 + 2^1 means that x = 1.
(It has been 65 years since I was in 10th grade, so my math may be a bit fuzzy.)
Awesome! Solve means find all solutions. We showed that there are no other solutions
Can we use the straightly increasing or decreasing function before or after a certain values to solve this equation.
It is obvious that 1 is a solution. We can test the function around this value maybe. 😉😉😉😉😉😉
Nice! Can explain in future video the theory / rationale behind AM GM?
Thanks! Yes, sure
x = 1