Problem Solving

ur wrote:please help in solving

1.(2^X)(3^Y) = 288 where x and y are positive integers,then (2^x-1)(3^y-2) =


2.3^x-3^x-1+162,then X(X-1)=

Thanks

1) 288 = 2^5*3^2 so 2^5-1*3^2-2 = 2^4*3^0 = 2^4*1

2) 3^x-3^x-1+162,then X(X-1)=

Note I am assuming the above statement is 3^x-3^x-1=162,then X(X-1)=

162 =3^4(2)

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

So x-1 = 4

so x = 5

so 5(4) = 20 (this is the ans)