1. ## Solving Logarithmic Equations

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
???????????????????

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
???????????????????

2. Originally Posted by Fallen Ghost
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
???????????????????

combine like terms ...
x^2 + 3x - 10 = 0
factor the quadratic to find the solutions for x ... remember to check the solutions for domain issues.

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
???????????????????

no ... [(x-1)(x+2)]/(x+6) = 1 , because

log 5 (x-1) + log 5 (x-2) - log 5 (x+6) = log 5 (1)

help any?

3. Originally Posted by Fallen Ghost
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?)

???????????????????

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 $\displaystyle {\color{red}\log_5 \frac{(x-1)(x+2)}{x+6} = 0 \Rightarrow \frac{(x-1)(x+2)}{x+6} = 5^0 = 1}$.

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.

???????????????????
..

4. Thanks a lot for the help, but both of you used the wrong expression in the second question because it was:

log 5 (x-1) + log 5 (x-2) - log 5 (x+6) = 0
not
log 5 (x-1) + log 5 (x+2) - log 5 (x+6) = 0

5. Originally Posted by Fallen Ghost
Thanks a lot for the help, but both of you used the wrong expression in the second question because it was:

log 5 (x-1) + log 5 (x-2) - log 5 (x+6) = 0
not
log 5 (x-1) + log 5 (x+2) - log 5 (x+6) = 0
And that would be because we both copied from your first line of working.

The mistake does not affect how we have showed you to solve this equation. I'm sure you can make the necessary correction.

6. That's true, sorry for my mistake and thank you both again for the wonderful help.