SOLUTION: Hi, I am having a problem solving: log(z^2-25)-log(z+5)=log7 The farthest I got was to re-write the problem to: log(z^2-25)(z+5)=7

Question 172047: Hi,
I am having a problem solving:
log(z^2-25)-log(z+5)=log7
The farthest I got was to re-write the problem to: log(z^2-25)(z+5)=7
Found 2 solutions by solver91311, Alan3354:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Close, but no cigar this time.

Remember the rules:

The sum of the logs is the log of the PRODUCT.

The difference of the logs is the log of the QUOTIENT.

So you should have written:

From here you should be able to recognize that so

Furthermore, if and only if , so:

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
log(z^2-25)-log(z+5)=log7
The farthest I got was to re-write the problem to: log(z^2-25)(z+5)=7
-------------
log(z^2-25)-log(z+5)=log7
log((z-5)*(z+5)) - log(z+5) = log(7) Factor z^2-25
log(z-5) + log(z+5) - log(z+5) = log(7) Add logs when multiplying
log(z-5) = log(7)
z-5 = 7
z = 12

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