Solving for x:
log_{4}(x+3)+log_{4}(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
Solving for x:
log_{4}(x+3)+log_{4}(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
**EDIT For the record, the OP originally posted this: $\displaystyle \log_4(x+3)=\log_4(x-3)=2$, which is what this response is based on **
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Do you actually mean this: one problem is $\displaystyle \log_4(x+3) = 2$, and a second problem is $\displaystyle \log_4(x-3) = 2$? Because there is no way that $\displaystyle \log_4(x-3) = \log_4 (x+3)$Assuming this is what you mean, then to solve for x remember the definition of a log: if $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$. So for the first problem you have $\displaystyle 4^2 = x+3$ so x = 13. How do you get 5?
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.
My apologies. I just found this site and posted in the wrong place. Perhaps a mod can use it. I'm actually in a college algebra class. I fell a little behind due to taking multiple classes and a full work schedule and Im playing catch up now. Its not easy being an older guy that went back to college.
So, looking at x = -5 and plugging it in then that means you something that isn't true. I think I understand it better now that I have written it out again. Correct me if Im wrong. THANK YOU for clearing this up.