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Logarithms anyone?

OK I am completely lost :eek: on this log equation:

log(3-x) + log(3+x) = log5

Does anyone get an answer of ±2? If so HOW did you do it?
 
Keep in mind that log(a) + log(b) = log(ab).

--J
 
Exactly, so this is what I did:

log(x^2 - 9) = log5

(x^2 -14) = 0

not even close to what I'm supposed to get, no idea how to factor what I just got!
 
How'd you get from

log(x^2 - 9) = log5

to

(x^2 -14) = 0

Are you sure you can do that?

--J
 

ShawnD

Science Advisor
658
1
ms. confused said:
Exactly, so this is what I did:

log(x^2 - 9) = log5

(x^2 -14) = 0
First of all, I think you expanded that wrong.
Secondly, that's not how you "undo" a log.

log(3-x) + log(3+x) = log5
log[(3-x)(3+x)] = log5
log(9 - x^2) = log5
9 - x^2 = 5
4 = x^2
x = +- 2
 
So, basically I shouldn't have changed (3-x) to (x-3) or (3+x) to (x+3)?
 
36
0
there is another way to solve this problem, using the following formulas.
[tex]log(\alpha)+log(\beta)=log(\alpha\cdot\beta)[/tex]
[tex]log(\alpha)-log(\beta)=log(\frac{\alpha}{\beta})[/tex]

about your second question :
[tex](x+3)=(3+x)[/tex]
[tex](x-3)\not=(3-x)[/tex]
 

ek

180
0
ShawnD said:
Secondly, that's not how you "undo" a log.
My high school math teacher used to call it "dropping logs". I always found that funny.

:tongue:
 

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