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

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
Close, but no cigar this time.

log%28%28z%5E2-25%29%29-log%28%28z%2B5%29%29=log%28%287%29%29

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:

log%28%28%28z%5E2-25%29%2F%28z%2B5%29%29%29=log%28%287%29%29

From here you should be able to recognize that z%5E2-25=%28z%2B5%29%28z-5%29 so log%28%28%28z%5E2-25%29%2F%28z%2B5%29%29%29=+log%28%28z-5%29%29+=+log%28%287%29%29

Furthermore, log%28b%2C%28a%29%29=log%28b%2C%28c%29%29 if and only if a=c, so:

z-5=7

z=12

Answer by Alan3354(69443) About Me  (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