Solving Logarithmic Equations: Log(3-x) + Log(3+x) = Log5

[ms. confused]

OK I am completely lost 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?

 

[Justin Lazear]

Keep in mind that log(a) + log(b) = log(ab).

--J

 

[ms. confused]

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!

 

[Justin Lazear]

How'd you get from

log(x^2 - 9) = log5

to

(x^2 -14) = 0

Are you sure you can do that?

--J

 

[ShawnD]

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

 

[ms. confused]

So, basically I shouldn't have changed (3-x) to (x-3) or (3+x) to (x+3)?

 

[boaz]

there is another way to solve this problem, using the following formulas.
$$log(\alpha)+log(\beta)=log(\alpha\cdot\beta)$$
$$log(\alpha)-log(\beta)=log(\frac{\alpha}{\beta})$$

about your second question :
$$(x+3)=(3+x)$$
$$(x-3)\not=(3-x)$$

 

[ek]

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: