SOLUTION: what the,,, log base 3 ( x + 16 )= 1 + log base 3 ( 2 - X )

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Question 30199: what the,,,
log base 3 ( x + 16 )= 1 + log base 3 ( 2 - X )
Found 2 solutions by sdmmadam@yahoo.com, Fermat:
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
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)



Answer by Fermat(136) About Me  (Show Source):
You can put this solution on YOUR website!
log_3(3) = 1
so we can write,
log_3(x+16) = log_3(3) + log_3(2-x)
log_3(x+16) = log_3(3*(2-x)) = log_3(6-3x)
log_3(x+16) = log_3(6-3x)
x+16 = 6-3x
4x = -10
x = -2.5
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