This method isn’t analytical. For instance, it doesn’t answer if this equation has only one solution. For the similar equations like that but with different coefficients it will not work.
Take a guess and assume that like 32, the answer is a power of two. Use logarithms in base 2 because log(2, 2^x) = x. and log(2, x^32) = 32 log(2, x). Or, x = 32 log(2, x). So x / log(2,x) = 32. This can be solved by inspection, again beginning with powers of 2: x is divided by x as some power of 2, Trying a few values of x we see that log(2,256) = 8 and 256/8 = 32.
Это не совсем подбор. Методология и тактика тут есть. Подбор был сделан под конец, что подпортило эффект. Подбора можно было и не делать. Но понадобилась бы вычислит машина. И снова была б критика ))
I use the Lambert W function to resolve x^32=2^x. «ln» both sides : 32 lnx= x ln2. So lnx/x = ln2/32 the same as (x^-1) lnx = ln2/32. x^-1 = e(ln(x^-1)) and lnx = -ln(x^-1). So we have : e(lnx^-1) x ln(x^-1)= -ln2/32. So lnx^-1 = W(-ln2/32). «e» both sides : x = 1/(e(W(-ln2/32))
Если условие найти натуральный корень, то решение класс, если количество корней, то по графику видно два и третий, который 256, его надо догадаться, за интересное уравнение спасибо!
Согласен, если условие - найти натуральный корень, то решение отлично подходит. Но по графику действительно видно два корня, а третий, который равен 256, нужно догадаться. Спасибо за интересное замечание и внимание к уравнению!
Если расписать левую и правую часть как функции, то за счёт чётности, функция f(x)=X^32 будет пересекать функцию g(x)=2^x в двух точках: одно решение потеряно. P. S. ради интереса забил в Desmosе эти 2 функции и абсциссы их пересечений равны 1.022 и -0.979. Вот так вот интересно.
@@Swamisonic18 Most likely, this is due to the fact that the polynomial of the nth degree has at most n roots. In our case, this is a polynomial of the 32nd degree on the right side, which essentially limits the number of roots of this equation. That is, in this equation it has already been proved that there are at least 3 real roots, and all the others are complex. (Unless we find other real roots)
The difference between \(2^{1.0224}\) and \((1.0224)^{32}\) is approximately \(-0.00044\). This small discrepancy indicates that \(1.0224\) is a very close approximation to the solution of the equation \(2^x = x^{32}\). ChatGPT
If you need to resort to manipulating in the way that you did, I think switching to logarithms is far easier. You simply have to find a multiple of 32 that includes as a factor a log to the base of 2 that generates itself....namely 256, which is 32 times log to the base 2 of, you guessed it, 256. If the kids are in the Olympiad selection process, that kind of arithmetic via logs is child's play to them.
Using logarithmic fuction is the true solution for this particular sum...because it is not a solution unless you are able to solve all the similar types of sums using that method, irrespective of the base and the exponent...for example if it was 31 instead of 32...using this method would have fetched us nothing at all.
x^(1/x) is a monotone decreasing function if x>e, then the equation x^(1/x)=2^(1/32) has at most one real solution. x=256 works, so no other real soultion.
Using a graphic calculator, I checked the points of contact between the two functions other than x=256. After that, based on the two coordinate values, I was able to plug more specific numbers into the given equation and found two values that were close to the correct answer. x=-0.979016935 x=1.0223929402
But Why we don’t do any not equivalent transformation? Or do? To loose another solution in real numbers? Or there is negative solutions have been lost?
Take log two of both sides then you get X equals log base two of x multiplied by 32 other words what number when you take it’s logarithm it does the same thing as dividing it by 32. You can make educated guesses until you stubble upon 256.
Great approach! Taking the log base 2 of both sides definitely simplifies the equation to x = log base 2 of x multiplied by 32. As you mentioned, making educated guesses can eventually lead to finding that x = 256. Thanks for sharing this insightful method!
Es una solución para estos exponentes con otros exponentes no funciona, no es un método solo es una solución, con logaritmos es más fácil resolver este tipo de ecuaciones.
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@@jeanpierre7971 yes, there's no guarantee for a unique solution If it's bijective, it has a unique sol What I meant is that x -> x^(1/x) does not need to be bijective to make one solution works as long as there is a verification at the end
Given that this is a competition question posed to us, there is no need to prove anything. All that is needed is determining a solution, not proving that it is a unique solution.
