Question

If X ka power Y+Z is equal to 1,Y ka power X+Z is equal to 1024 and Z ka power X+Y is equal to729(X,Y,Z are natural number)then what is the value of (Z+1)ka power Y+X+1

Answer 1

Answer:3.862

Step-by-step explanation:

Given that,

X^(y+z)=1,. Y^(x+z)=1024. And Z^(x+y)=729

To find: (Z+1)^(y+x+1)=?

Taking log on both sides we get

(Y+Z)log x=log 1

Or,(y+z)log x=0.........(1)

Again,

(X+Z)log y= log 1024

Or, (x+z) log y=3.010............(2)

Also,

(X+Y)log z= log 729

Or,(x+y)log z=2.862................(3)

Now,log z=2.862/(x+y)

Log (z+1)=2.862+1/x+y+1

Or,log (z+1)=3.862/(x+y+1)

Or,(z+1)^(x+y+1)=3.862