solve the equeation

Got this one from goiit.. (Nice that i went there and browsed a couple of posts :)

21 = x^y + y^x
Find the values of x and y . (Restrict to the integral solutions only)

Discuss the other cases too...

9 Answers

11
Gone.. ·

20 and 1
can be a solution.

62
Lokesh Verma ·

Is that the only solution.. and can we think for it in a more general way?

3
msp ·

is there ne soln not by observation :P,bcos the function xy and its inverse was complicated to me.

62
Lokesh Verma ·

No I dont think there is another solution not by observation..

BUt by graphs ther is a way if you can think nicely

11
Gone.. ·

bhaiya only integral solutions is asked ..dont think there is any other integral solution.

23
qwerty ·

there is one more of dis type .... xy + yx = 8

1
rickde ·

x^y<21 x≠1 y≠1
so permissible values of
x y
2 2,3,4
3 2,
4 2

we see none of these satisfy

hence 20,1 and 1,20 are only soln

for x^y+y^x=8

proceeding similarly we get
1,8 8,1 and 2,2

11
Devil ·

Alternately we can do something of this kind ....
(1,20) & (20,1) are the obvious ones....Other than those....if we have both x\ge 1\\ & y\ge 1\\ then :-

y^x\ge 4 and x^y\ge x^2 so 21\ge x^2+4
Meaning probable x can be (2,3,4) Easy to see none satisfy the eqn....

how ?
Like this :-
Suppose x=4, then y^4-21=-4^y ,meaning y can be at max 2.....checking like this, we confirm no other solutions.....

BTW check the following link!

11
Devil ·

http://www.goiit.com/posts/list/algebra-without-binomial-98498.htm

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