Functions are confusing. Why is log(x - 2) + log(x + 2) not the same function as log(x^2 - 4)?
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topsquark
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What's the domain of log(x-2) + log(x+2)? It's x > 2.
What's the domain of [imath]log(x^2-4)[/imath]? It's [imath]x < -2 \bigcup 2 < x[/imath].
In the first expression, it's only the first term that determines the domain.
Think of it this way: What is log(-4) + log(-4)? It doesn't exist. But log(-4) + log(-4) = log(16), which does exist. It depends on how you write it.
In the end, if you use the complex numbers for your base set, both of these expressions do have the same domain: [imath]x \neq -2, 2[/imath].
-Dan
Harry_the_cat
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When you are asked to find the domain of a function, you are basically asked to find which x-values "work" in the function.
The function log (x) is only defined when x>0.
In f(x) = log(x-2) + log(x+2), you need to make sure that both (x-2)>0 AND (x+2)>0.
(x-2)>0 implies x>2
(x+2)>0 implies x>-2
So if x>2 AND x>-2, then x>2. So the domain of (1) is 2<x<oo or (2, oo) depending on which notation you use.
In f(x) = log(x^2 - 4), you need to make sure that x^2 - 4 > 0
x^2 - 4 > 0 implies that x^2 >4. So either x<-2 OR x>2. So the domain of (2) is (-oo, -2) U (2, oo).
I suggest you draw the graphs of both on a graphics calculator or Desmos, and you will see the graphs are not the same.
The function log (x) is only defined when x>0.
In f(x) = log(x-2) + log(x+2), you need to make sure that both (x-2)>0 AND (x+2)>0.
(x-2)>0 implies x>2
(x+2)>0 implies x>-2
So if x>2 AND x>-2, then x>2. So the domain of (1) is 2<x<oo or (2, oo) depending on which notation you use.
In f(x) = log(x^2 - 4), you need to make sure that x^2 - 4 > 0
x^2 - 4 > 0 implies that x^2 >4. So either x<-2 OR x>2. So the domain of (2) is (-oo, -2) U (2, oo).
I suggest you draw the graphs of both on a graphics calculator or Desmos, and you will see the graphs are not the same.
blamocur
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f(x) in 1. is not defined for x=-3, but it is in 2..
Dr.Peterson
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What's happening here is that simplifying (or, more generally, rewriting using properties of the log) can change the domain, so that the result is technically a different function.
In particular, the property that log(ab) = log(a) + log(b) is valid only when both sides are defined, The LHS is defined for ab > 0, the RHS only when both a and b are greater than 0 (and not when they are both negative). So you must always consider the domains separately, and not just assume they are the same.
I appreciate what you are all saying but I still find this a bit weird. What to me are equivalent functions not having the same domain still confuses me.
Thanks for your help. I’ll consider your replies for a while and hopefully a light bulb will come on at some stage.
Thanks for your help. I’ll consider your replies for a while and hopefully a light bulb will come on at some stage.
blamocur
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Strictly speaking, the functions are not equivalent
Steven G
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Here is what is going on--basically what Dr Peterson said.
Log (AB) = Log A + log B is NOT always true. It is true only if AB>0, A>0 & B>0
Log (A/B) = Log A - Log B is NOT always true. It is true only if A/B>0, A>0 and B>0
Log Ar = rLog A is NOT always true. It is true only if A>0.
The above three equations are NOT equivalent unless the restraints listed above are true.
Ex: Log(-2)2 is a valid amount as it equals Log(4). However, 2Log(-2) is not valid as Log(-2) makes no sense.
log(-4) is undefined. log(4) is defined.
log(-4) + log(-4) = 2*log(-4)
log(16) = log(4^2) = 2*log(4) = log(4) + log(4)
2*log(-4) = 2*log(4) ? ==>
log(-4) = log(4) ?
From the top line, the left-hand side is undefined, but the right-hand side is defined.
log(A*B) = log(A) + log(B) for appropriate A and B values.