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?
logarithms wolfram-alpha
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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?
logarithms wolfram-alpha
Likely itâÂÂs assuming that $y$ is complex.
â amd
Aug 29 at 9:32
What do you expected e.g. for $y=-3?$
â 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 $y in mathbb R$.
â xbh
Aug 29 at 9:36
1
Or when $y=-1$ ? (oops - xbh just said that !)
â 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.
â awkward
Aug 29 at 12:23
add a comment |Â
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?
logarithms wolfram-alpha
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?
logarithms wolfram-alpha
asked Aug 29 at 9:28
Dan Hoynoski
1527
1527
Likely itâÂÂs assuming that $y$ is complex.
â amd
Aug 29 at 9:32
What do you expected e.g. for $y=-3?$
â 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 $y in mathbb R$.
â xbh
Aug 29 at 9:36
1
Or when $y=-1$ ? (oops - xbh just said that !)
â 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.
â awkward
Aug 29 at 12:23
add a comment |Â
Likely itâÂÂs assuming that $y$ is complex.
â amd
Aug 29 at 9:32
What do you expected e.g. for $y=-3?$
â 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 $y in mathbb R$.
â xbh
Aug 29 at 9:36
1
Or when $y=-1$ ? (oops - xbh just said that !)
â 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.
â awkward
Aug 29 at 12:23
Likely itâÂÂs assuming that $y$ is complex.
â amd
Aug 29 at 9:32
Likely itâÂÂs assuming that $y$ is complex.
â amd
Aug 29 at 9:32
What do you expected e.g. for $y=-3?$
â gammatester
Aug 29 at 9:34
What do you expected e.g. for $y=-3?$
â 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 $y in mathbb R$.
â 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 $y in mathbb R$.
â xbh
Aug 29 at 9:36
1
1
Or when $y=-1$ ? (oops - xbh just said that !)
â gandalf61
Aug 29 at 9:36
Or when $y=-1$ ? (oops - xbh just said that !)
â 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.
â 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.
â awkward
Aug 29 at 12:23
add a comment |Â
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 !
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?
â Dan Hoynoski
Aug 29 at 10:17
add a comment |Â
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 !
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?
â Dan Hoynoski
Aug 29 at 10:17
add a comment |Â
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 !
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?
â Dan Hoynoski
Aug 29 at 10:17
add a comment |Â
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 !
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 !
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?
â Dan Hoynoski
Aug 29 at 10:17
add a comment |Â
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?
â 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?
â 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?
â Dan Hoynoski
Aug 29 at 10:17
add a comment |Â
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Likely itâÂÂs assuming that $y$ is complex.
â amd
Aug 29 at 9:32
What do you expected e.g. for $y=-3?$
â 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 $y in mathbb R$.
â xbh
Aug 29 at 9:36
1
Or when $y=-1$ ? (oops - xbh just said that !)
â 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.
â awkward
Aug 29 at 12:23