I Solved An Exponential Equation from Romania

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[altosrule]

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) + + (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.

[weylguy]

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!

[alextang]

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. 😉😉😉😉😉😉

[scottleung]

I also got x=1 as the only real solution.

[alexestefan]

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!

[monkeblazer]

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)

[error__brain_deleted]

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.)

[ilanbar]

Nice! Can explain in future video the theory / rationale behind AM GM?