Hey guys, I'm having trouble with a couple of grade 12 math questions. A friend of mine told me about this forum, and spoke highly of the positive response to all of his posts. I have been attempting these questions for so long but I can't seem to find a solution. These question are homework material. I know the answers (well if the back of the book is correct) but i need to understand the concepts behind the work.
These are my questions:
1. log a (x+2) + log a (x-1) = log a (8-2x)
Attempt:
***Well we know that when log a M = log a N then M = N***
Since the base of the log's are the same we can drop the log a
According to product exponent law we know that: (a^x) (a^y) = (a^x+y)
(x+2) (x-1) = (8-2x)
x^2 -x +2x - 2 - 8 + 2x = 0
Mr F says: Therefore x^2 + 3x - 10 = 0 => (x + 5)(x - 2) = 0 .... One of these solutions is invalid ..... (why?)
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2. log 5 (x-1) + log 5 (x-2) - log 5 (x+6) = 0
Attempt:
Since the base of the log's are the same we can drop the log 5
According to product exponent law we know that: (a^x) (a^y) = (a^x+y)
According to quotient exponent law we know that: (a^x)/(a^y) = (a^x-y)
( (x-1) (x+2) / (x+6) ) = 0
Mr F says: Therefore (x+2)}{x+6} = 0 \Rightarrow \frac{(x-1)(x+2)}{x+6} = 5^0 = 1})
.
Therefore (x - 1)(x + 2) = x + 6. Now expand the left hand side and simplify the equation into the form quadratic = 0. Then solve the quadratic for x. You only want solutions that satisfy the original equation.
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Solving Logarithmic Equations
Originally Posted by Fallen Ghost 
