Wolfram Alpha Ignoring Log Rules?

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Why does Wolfram Alpha say that $underbraceln(y(y^2-4))_textEQ1 neq underbraceln(y) + ln(y-2) + ln(y+2)_textEQ2$ (seen here)?



The Produce Rule of Logarithms states that:
$$log_b(M*N)=log_bM + log_bN$$



Therefore,
$$ beginalign ln(y) + ln(y-2) + ln(y+2)& = \ ln(y) + ln((y-2)(y+2)) & = \ ln(y) + ln(y^2-4) & = ln(y(y^2 - 4)) = mathbfEQ1 checkmark endalign $$



However, Wolfram Alpha is saying that EQ1 "is not always equal to" EQ2. The graph Wolfram provides feels problematic, because it is showing two dramatically different curves when in reality both sizes are the same curve. What am I missing here?







share|cite|improve this question




















  • Likely it’s assuming that $y$ is complex.
    – amd
    Aug 29 at 9:32










  • What do you expected e.g. for $y=-3?$
    – gammatester
    Aug 29 at 9:34






  • 1




    For $y = -1$, the LHS is well defined, but for the RHS, none of them are valid when considering $y in mathbb R$.
    – xbh
    Aug 29 at 9:36






  • 1




    Or when $y=-1$ ? (oops - xbh just said that !)
    – gandalf61
    Aug 29 at 9:36











  • In Mathematica, "Simplify[Log[y (y^2 - 4)] - Log[y] - Log[y - 2] - Log[y + 2], Assumptions -> y > 2]" works, but I don't know how to do this in Alpha.
    – awkward
    Aug 29 at 12:23














up vote
0
down vote

favorite












Why does Wolfram Alpha say that $underbraceln(y(y^2-4))_textEQ1 neq underbraceln(y) + ln(y-2) + ln(y+2)_textEQ2$ (seen here)?



The Produce Rule of Logarithms states that:
$$log_b(M*N)=log_bM + log_bN$$



Therefore,
$$ beginalign ln(y) + ln(y-2) + ln(y+2)& = \ ln(y) + ln((y-2)(y+2)) & = \ ln(y) + ln(y^2-4) & = ln(y(y^2 - 4)) = mathbfEQ1 checkmark endalign $$



However, Wolfram Alpha is saying that EQ1 "is not always equal to" EQ2. The graph Wolfram provides feels problematic, because it is showing two dramatically different curves when in reality both sizes are the same curve. What am I missing here?







share|cite|improve this question




















  • Likely it’s assuming that $y$ is complex.
    – amd
    Aug 29 at 9:32










  • What do you expected e.g. for $y=-3?$
    – gammatester
    Aug 29 at 9:34






  • 1




    For $y = -1$, the LHS is well defined, but for the RHS, none of them are valid when considering $y in mathbb R$.
    – xbh
    Aug 29 at 9:36






  • 1




    Or when $y=-1$ ? (oops - xbh just said that !)
    – gandalf61
    Aug 29 at 9:36











  • In Mathematica, "Simplify[Log[y (y^2 - 4)] - Log[y] - Log[y - 2] - Log[y + 2], Assumptions -> y > 2]" works, but I don't know how to do this in Alpha.
    – awkward
    Aug 29 at 12:23












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Why does Wolfram Alpha say that $underbraceln(y(y^2-4))_textEQ1 neq underbraceln(y) + ln(y-2) + ln(y+2)_textEQ2$ (seen here)?



The Produce Rule of Logarithms states that:
$$log_b(M*N)=log_bM + log_bN$$



Therefore,
$$ beginalign ln(y) + ln(y-2) + ln(y+2)& = \ ln(y) + ln((y-2)(y+2)) & = \ ln(y) + ln(y^2-4) & = ln(y(y^2 - 4)) = mathbfEQ1 checkmark endalign $$



However, Wolfram Alpha is saying that EQ1 "is not always equal to" EQ2. The graph Wolfram provides feels problematic, because it is showing two dramatically different curves when in reality both sizes are the same curve. What am I missing here?







share|cite|improve this question












Why does Wolfram Alpha say that $underbraceln(y(y^2-4))_textEQ1 neq underbraceln(y) + ln(y-2) + ln(y+2)_textEQ2$ (seen here)?



The Produce Rule of Logarithms states that:
$$log_b(M*N)=log_bM + log_bN$$



Therefore,
$$ beginalign ln(y) + ln(y-2) + ln(y+2)& = \ ln(y) + ln((y-2)(y+2)) & = \ ln(y) + ln(y^2-4) & = ln(y(y^2 - 4)) = mathbfEQ1 checkmark endalign $$



However, Wolfram Alpha is saying that EQ1 "is not always equal to" EQ2. The graph Wolfram provides feels problematic, because it is showing two dramatically different curves when in reality both sizes are the same curve. What am I missing here?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 29 at 9:28









Dan Hoynoski

1527




1527











  • Likely it’s assuming that $y$ is complex.
    – amd
    Aug 29 at 9:32










  • What do you expected e.g. for $y=-3?$
    – gammatester
    Aug 29 at 9:34






  • 1




    For $y = -1$, the LHS is well defined, but for the RHS, none of them are valid when considering $y in mathbb R$.
    – xbh
    Aug 29 at 9:36






  • 1




    Or when $y=-1$ ? (oops - xbh just said that !)
    – gandalf61
    Aug 29 at 9:36











  • In Mathematica, "Simplify[Log[y (y^2 - 4)] - Log[y] - Log[y - 2] - Log[y + 2], Assumptions -> y > 2]" works, but I don't know how to do this in Alpha.
    – awkward
    Aug 29 at 12:23
















  • Likely it’s assuming that $y$ is complex.
    – amd
    Aug 29 at 9:32










  • What do you expected e.g. for $y=-3?$
    – gammatester
    Aug 29 at 9:34






  • 1




    For $y = -1$, the LHS is well defined, but for the RHS, none of them are valid when considering $y in mathbb R$.
    – xbh
    Aug 29 at 9:36






  • 1




    Or when $y=-1$ ? (oops - xbh just said that !)
    – gandalf61
    Aug 29 at 9:36











  • In Mathematica, "Simplify[Log[y (y^2 - 4)] - Log[y] - Log[y - 2] - Log[y + 2], Assumptions -> y > 2]" works, but I don't know how to do this in Alpha.
    – awkward
    Aug 29 at 12:23















Likely it’s assuming that $y$ is complex.
– amd
Aug 29 at 9:32




Likely it’s assuming that $y$ is complex.
– amd
Aug 29 at 9:32












What do you expected e.g. for $y=-3?$
– gammatester
Aug 29 at 9:34




What do you expected e.g. for $y=-3?$
– gammatester
Aug 29 at 9:34




1




1




For $y = -1$, the LHS is well defined, but for the RHS, none of them are valid when considering $y in mathbb R$.
– xbh
Aug 29 at 9:36




For $y = -1$, the LHS is well defined, but for the RHS, none of them are valid when considering $y in mathbb R$.
– xbh
Aug 29 at 9:36




1




1




Or when $y=-1$ ? (oops - xbh just said that !)
– gandalf61
Aug 29 at 9:36





Or when $y=-1$ ? (oops - xbh just said that !)
– gandalf61
Aug 29 at 9:36













In Mathematica, "Simplify[Log[y (y^2 - 4)] - Log[y] - Log[y - 2] - Log[y + 2], Assumptions -> y > 2]" works, but I don't know how to do this in Alpha.
– awkward
Aug 29 at 12:23




In Mathematica, "Simplify[Log[y (y^2 - 4)] - Log[y] - Log[y - 2] - Log[y + 2], Assumptions -> y > 2]" works, but I don't know how to do this in Alpha.
– awkward
Aug 29 at 12:23










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










For example: $ln(y(y^2-4))$ is defined for $y=-1$, but $ln(y) + ln(y-2) + ln(y+2)$ is not defined for $y=-1$.



The rule $log_b(M*N)=log_bM + log_bN$ is only valid if both $M$ and $N$ are positive !






share|cite|improve this answer




















  • Quick question, if I was to encapsulate each interior with absolute values (i.e. $ln|y(y^2-4)| stackrel?= ln|y| + ln|y-2| + ln|y+2|$), why is it still not always equal to each other?
    – Dan Hoynoski
    Aug 29 at 10:17











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










For example: $ln(y(y^2-4))$ is defined for $y=-1$, but $ln(y) + ln(y-2) + ln(y+2)$ is not defined for $y=-1$.



The rule $log_b(M*N)=log_bM + log_bN$ is only valid if both $M$ and $N$ are positive !






share|cite|improve this answer




















  • Quick question, if I was to encapsulate each interior with absolute values (i.e. $ln|y(y^2-4)| stackrel?= ln|y| + ln|y-2| + ln|y+2|$), why is it still not always equal to each other?
    – Dan Hoynoski
    Aug 29 at 10:17















up vote
2
down vote



accepted










For example: $ln(y(y^2-4))$ is defined for $y=-1$, but $ln(y) + ln(y-2) + ln(y+2)$ is not defined for $y=-1$.



The rule $log_b(M*N)=log_bM + log_bN$ is only valid if both $M$ and $N$ are positive !






share|cite|improve this answer




















  • Quick question, if I was to encapsulate each interior with absolute values (i.e. $ln|y(y^2-4)| stackrel?= ln|y| + ln|y-2| + ln|y+2|$), why is it still not always equal to each other?
    – Dan Hoynoski
    Aug 29 at 10:17













up vote
2
down vote



accepted







up vote
2
down vote



accepted






For example: $ln(y(y^2-4))$ is defined for $y=-1$, but $ln(y) + ln(y-2) + ln(y+2)$ is not defined for $y=-1$.



The rule $log_b(M*N)=log_bM + log_bN$ is only valid if both $M$ and $N$ are positive !






share|cite|improve this answer












For example: $ln(y(y^2-4))$ is defined for $y=-1$, but $ln(y) + ln(y-2) + ln(y+2)$ is not defined for $y=-1$.



The rule $log_b(M*N)=log_bM + log_bN$ is only valid if both $M$ and $N$ are positive !







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 29 at 9:35









Fred

38.2k1238




38.2k1238











  • Quick question, if I was to encapsulate each interior with absolute values (i.e. $ln|y(y^2-4)| stackrel?= ln|y| + ln|y-2| + ln|y+2|$), why is it still not always equal to each other?
    – Dan Hoynoski
    Aug 29 at 10:17

















  • Quick question, if I was to encapsulate each interior with absolute values (i.e. $ln|y(y^2-4)| stackrel?= ln|y| + ln|y-2| + ln|y+2|$), why is it still not always equal to each other?
    – Dan Hoynoski
    Aug 29 at 10:17
















Quick question, if I was to encapsulate each interior with absolute values (i.e. $ln|y(y^2-4)| stackrel?= ln|y| + ln|y-2| + ln|y+2|$), why is it still not always equal to each other?
– Dan Hoynoski
Aug 29 at 10:17





Quick question, if I was to encapsulate each interior with absolute values (i.e. $ln|y(y^2-4)| stackrel?= ln|y| + ln|y-2| + ln|y+2|$), why is it still not always equal to each other?
– Dan Hoynoski
Aug 29 at 10:17


















 

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