hii

 

well, see let us do it like this .. for each term we find out the condition on x and then try to get intersection of all ..

 

 x - 3 > 0     .. (1)

 x - 1 > 0     .. (2)

 

 also, - log_x-3(x-1) > 0  ( since it is in the sq rt sign )

 

Now, from (1) and (2) we know that x > 3

 

we want log_x-3(x-1) <  0 , so now x - 1 > 2 , so x - 3 < 1 is the only way left to make it less than 0

 

thus we get x < 4   ... (3)

 

combining (1) and (3) we get,

 

                                     3 < x < 4          ... (4)

 

Now the term -x²+2x+8 > 0

 

or,         9 - (x - 1)² > 0

 

which means 3 > x - 1 > -3

 

or,    4 > x > - 2                                     .. (5)

 

From (4) and (5) we get

 

 3 < x < 4    .. Ans

 

 

cheers