Question

If, 2^x+2^y+2^z=2336.find x, y, z

Answer 1

X, Y and Z is equivalent to 5, 8 and 11
(2^5)+(2^8)+(2^11)=2336
(2^5=32)+(2^8=256)+(2^11=2048)=2336

Answer 2

Answer:

X = 5 , Y = 8 , Z = 11

Step-by-step explanation:

$2^{x} + 2^{y} + 2^{z} = 2336\\2^{x}(1 + 2^{y-x} + 2^{z-x}) = 2336\\2^{x}(1 +2^{y-x} + 2^{z-x} ) = 2^{5} * 73$

On comparing we can say x = 5 , hence :

$1 + 2^{y-x} + 2^{z-x} = 73\\2^{y-x} + 2^{z-x} = 72\\2^{y-x}(1 + 2^{z-x-y+x}) = 2^{3}*3^{2}$

On comparing we can again say y-x = 3 , hence :

$1 + 2^{z-y} = 9\\2^{z-y} = 2^3$

On comparing again we can say z-y = 3

Now we can easily solve these 3 equations :

x-5 = 0

y-x = 3

z-y = 3

After solving these 3 equations we will find that the value of <x,y,z>  is <5,8,11>

Hope This Might Help Someone !