3:08 directly take log both sides naa (x - y)(x / y) = 3 log2 x - y = 3 ......(i) x / y = 2 x = 2y put x in (i), 2y - y = 3 y = 3 now put y in (i) x / 3 = 2 x = 6
@@superacademy247 x=y=0 is not necessarily a solution! We have lim_{a-> 0} a^a = lim_{a-> 0} exp(a ln a) = exp(0) = 1. So it is not unreasonable to define 0^0 = 1. Inserted in the original equation we obtain 1 = 8, which is false.
3:08 directly take log both sides naa
(x - y)(x / y) = 3 log2
x - y = 3 ......(i)
x / y = 2
x = 2y
put x in (i),
2y - y = 3
y = 3
now put y in (i)
x / 3 = 2
x = 6
👍👍♥👍❤♥
More general solution required to show that
no other integer solution is possible .
(x^x)y^y = 8(x^y)y^x
[x^(x - y)]y^(y - x) = 8
(x/y)^(x - y) = 2³
x - y = 3
x/y = 2 => x = 2y
2y - y = 3 => *y = 3* => *x = 6*
Нельзя делить на какие то переменные, не убедившись что они не равны 0. Потому что х=0 или у=0, вполне могут быть корнями уравнения
You forgot the solution x=y=0
Yeah. Zero is also a solution!
@@superacademy247 x=y=0 is not necessarily a solution! We have lim_{a-> 0} a^a = lim_{a-> 0} exp(a ln a) = exp(0) = 1. So it is not unreasonable to define 0^0 = 1. Inserted in the original equation we obtain 1 = 8, which is false.