Question

If 2^x-1=3^y and 4^x=3^y+1 then find x and y
Plzzzzzzzzzzzz it's urgent

Answer 1

Step-by-step explanation:

given:

2^x - 1=3^y....equ(i)

4^x=3^y+1

=>2^2x=2^x - 1+1 from equ(i)

=>2^2x-2^x=0

=>2^x(2^x-1)=0

=>2^x-1=0 [ as 2^x≠0]

=>2^x=2^0

=>x=0 [equating powers since base is equal]

now substituting x=0 in equ(i),

2^0-1=3^y

=>1-1=3^y

=>3^y=0

which is not possible!!!!

no thanks

Answer 2

Question:

If 2^x - 1 = 3^y and 4^x = 3^y + 1 then find the values of x and y.

Solution:

We have ;

2^x - 1 = 3^y ----------(1)

4^x = 3^y + 1 -----------(2)

Let,

a = 2^x and b = 3^y

Now, putting a = 2^x and b = 3^y in the given equations, we get;

=> 2^x - 1 = 3^y

=> a - 1 = b ---------(3)

Also;

=> 4^x = 3^y + 1

=> (2^2)^x = 3y + 1

=> 2^2x = 3y + 1

=> (2^x)^2 = 3y + 1

=> a^2 = b + 1

=> b = a^2 - 1 ----------(4)

Now,

From eq-(3) and eq-(4) , we have;

=> a - 1 = a^2 - 1

=> a^2 - a = 0

=> a(a - 1) = 0

=> a = 0 or 1

If a = 0 ,then;

=> 2^x = 0

{ This is not possible, as 2 is a positive number, and any power to a positive number can't be zero }

If a = 1 , then;

=> 2^x = 1

=> 2^x = 2^0

=> x = 0

Also,

Putting a = 1 , in eq-(4) , we get;

=> b = a^2 - 1

=> b = (1)^2 - 1

=> b = 1 - 1

=> b = 0

=> 3^y = 0

{ This is not possible, as 3 is a positive number, and any power to a positive number can't be zero }

Hence,

There doesn't exist any real value of x and y which satisfy the given system of equations.