Question 30199
log base 3 ( x + 16 )= 1 + log base 3 ( 2 - X )  ----(1)
understanding that the base is 3 we write the above equation for convenience as
log( x + 16 )= 1 + log( 2 - X )
log( x + 16 )- log( 2 - X )=1  (change side then change sign)
log[(x+16)/(2-x)] = 1 (using loga-logb =log(/b) of course all for the same base)
That is [(x+16)/(2-x)] = 3^1   
(using definition:log baseb(N) = p implies and is implied by N = b^p where N>0 )
Multiplying by (2-x) on both the sides
(x+16) = 3(2-x)----(2)
x+16 = 6 - 3x
x+3x = 6 -16
4x = -10
x=(-10)/4 = (-5/2)
Note: since x = -5/2, we must check 
whether (x+16) and (2-x) are positive for this value of x.
Yes. They are positive and hence we can accept the value x =-5/2
Let us now verify in the equivalent form (2)
LHS = (x+16) =(-5/2) + 16 = 27/2
And RHS = 6-3x = 6+15/2 = 27/2 = LHS
Therefore our value is correct
Answer: x = (-5/2)