# Log solve confusion

• November 30th 2012, 09:11 AM
Log solve confusion
Solving for x:

log4(x+3)+log4(x-3)=2

I'm doing something incorrect, I have gotten answers working through this as x=5 and x=5,-5

I'm trying to understand why, and if in the end I would need to enter as a correct solution x=5 OR x=5,-5
• November 30th 2012, 09:18 AM
ebaines
Re: Log solve confusion
**EDIT For the record, the OP originally posted this: $\log_4(x+3)=\log_4(x-3)=2$, which is what this response is based on **
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Do you actually mean this: one problem is $\log_4(x+3) = 2$, and a second problem is $\log_4(x-3) = 2$? Because there is no way that $\log_4(x-3) = \log_4 (x+3)$Assuming this is what you mean, then to solve for x remember the definition of a log: if $\log_a b = c$ then $a^c = b$. So for the first problem you have $4^2 = x+3$ so x = 13. How do you get 5?
• November 30th 2012, 09:22 AM
Plato
Re: Log solve confusion
Quote:

Originally Posted by gbrad88
Solving for x:l
og4(x+3)=log4(x-3)=2
I'm doing something incorrect, I have gotten answers working through this as x=5 and x=5,-5
I'm trying to understand why, and if in the end I would need to enter as a correct solution x=5 OR x=5,-5

I think that you mean $\log_4(x+3)+\log_4(x-3)=2$.

That gives $(x+3)(x-3)=4^2$ or $x^2-9=16$.

So $x=\pm 5~.$
• November 30th 2012, 09:35 AM
ebaines
Re: Log solve confusion
Quote:

Originally Posted by Plato
I think that you mean $\log_4(x+3)+\log_4(x-3)=2$.

Good detective work!

Quote:

Originally Posted by Plato
So $x=\pm 5~.$

But if x = -5 then that means you have $\log_4 (-2) + \log_4 (-8) = 2$, which isn't true. The answer is limited to x = +5.
• November 30th 2012, 09:43 AM
Re: Log solve confusion
ok, so my work that I did for:

Solving for x:

log4(x+3)+log4(x-3)=2

would be 5,-5 or just 5?

Sorry the response confused me?

From reading, it seems it SHOULD be just "5"?
• November 30th 2012, 09:53 AM
ebaines
Re: Log solve confusion
gbrad88 - since you asked this in a pre-university algebra forum, can I assume that you have not covered complex numbers in your math classes? If that's so, then the logarithm of negative numbers is not something to consider, and for now you should limit your answers such that (x+3) and (x-3) are postive numbers only.
• November 30th 2012, 10:00 AM