解方程：2ˣ=x³²，学霸的方法很少有人想得到！ 

x^32=2^x 令x=2^n，且n為正整數 令n=2^a，且a為正整數 x^32=2^x 2^n^32=2^2^n 因為2^32n=2^2^n 所以32n=2^n n=2^a a代1 2(2^5)=2^2矛盾❎ a代2 4(2^5)=2^2矛盾❎ a代3 8(2^5)=2^2正確✅ n=2^a x=2^n =2^3 =2^8 =8 =256 A:256

绘图是y=x 和 y=2^(x-5)两条曲线有两个交点就有两个解，一个是256，还有一个是0.0319438

聽過lambert W function 嗎

兩邊取2為底的對數就能解了

Thank you for explaining. (I am NOT Chinese.) I would like to confirm original problem's condition. If the condition is "x is integer", the solution is x=256. But, if the condition is "x is real number", there are other solutions. One is between 1 and 2, the other is between -1 and 0. (I don't know Chinese, so, I cannot get the information about that from this video. Therefore, I am asking about original problem's condition.)

2^0>0^32 2^(-1)1^32 2^x&x^32連續可微 可證x不具備唯一解

xlog2(2)=32log2(x)

我第一眼看出来的是有一个分数解……

畫個圖形就知道 不只一個解啊 x~1.02239、-0.97902

X=256

1.0224