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How to plot implicit function $log_10(xy-1) = log_10((1-x)(1-y))$
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I'm inspecting various function and try to plot their graph based on my observations and compare with graphing tools afterwards.
Given an implicit function $log_10(xy-1) = log_10((1-x)(1-y))$ how do i plot it?
This comes really hard to me since I have strong doubts about the domain of the function. I can't just reduce the $log$ since it comes with some constraints.
Basically what i think is that the above reduces to:
$$
|xy-1|=|(1-x)(1-y)|,
$$
since logarithm argument is greater than $0$ (at least for $xin mathbb R$).
But now how do i find the explicit form of $y=f(x)$ considering the constraints implied by the logarithm?
upd:
Here is how the graph looks, build based on the answers below:
Intersection of black an purple areas is the domain of the function, and red line is $y=2-x$. Those parts never cross hence no solution in reals is present. Based on that i conclude that the graph is just an empty plane
algebra-precalculus logarithms graphing-functions
asked Jul 27 at 15:43
roman
4341412
add a comment |Â
up vote
1
down vote
favorite
I'm inspecting various function and try to plot their graph based on my observations and compare with graphing tools afterwards.
Given an implicit function $log_10(xy-1) = log_10((1-x)(1-y))$ how do i plot it?
This comes really hard to me since I have strong doubts about the domain of the function. I can't just reduce the $log$ since it comes with some constraints.
Basically what i think is that the above reduces to:
$$
|xy-1|=|(1-x)(1-y)|,
$$
since logarithm argument is greater than $0$ (at least for $xin mathbb R$).
But now how do i find the explicit form of $y=f(x)$ considering the constraints implied by the logarithm?
upd:
Here is how the graph looks, build based on the answers below:
Intersection of black an purple areas is the domain of the function, and red line is $y=2-x$. Those parts never cross hence no solution in reals is present. Based on that i conclude that the graph is just an empty plane
algebra-precalculus logarithms graphing-functions
asked Jul 27 at 15:43
roman
4341412
Helps to note that $log_aX=log_a Yimplies X=Y$ (and conversely, if $X,Y>0$).
– lulu
Jul 27 at 15:45
Your formula involves no absolute value.
– Bernard
Jul 27 at 15:51
Keep in mind that injectivity for $log$ is essential to obtain the result.
– gimusi
Jul 27 at 16:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm inspecting various function and try to plot their graph based on my observations and compare with graphing tools afterwards.
Given an implicit function $log_10(xy-1) = log_10((1-x)(1-y))$ how do i plot it?
This comes really hard to me since I have strong doubts about the domain of the function. I can't just reduce the $log$ since it comes with some constraints.
Basically what i think is that the above reduces to:
$$
|xy-1|=|(1-x)(1-y)|,
$$
since logarithm argument is greater than $0$ (at least for $xin mathbb R$).
But now how do i find the explicit form of $y=f(x)$ considering the constraints implied by the logarithm?
upd:
Here is how the graph looks, build based on the answers below:
Intersection of black an purple areas is the domain of the function, and red line is $y=2-x$. Those parts never cross hence no solution in reals is present. Based on that i conclude that the graph is just an empty plane
algebra-precalculus logarithms graphing-functions
asked Jul 27 at 15:43
roman
4341412
I'm inspecting various function and try to plot their graph based on my observations and compare with graphing tools afterwards.
Given an implicit function $log_10(xy-1) = log_10((1-x)(1-y))$ how do i plot it?
This comes really hard to me since I have strong doubts about the domain of the function. I can't just reduce the $log$ since it comes with some constraints.
Basically what i think is that the above reduces to:
$$
|xy-1|=|(1-x)(1-y)|,
$$
since logarithm argument is greater than $0$ (at least for $xin mathbb R$).
But now how do i find the explicit form of $y=f(x)$ considering the constraints implied by the logarithm?
upd:
Here is how the graph looks, build based on the answers below:
Intersection of black an purple areas is the domain of the function, and red line is $y=2-x$. Those parts never cross hence no solution in reals is present. Based on that i conclude that the graph is just an empty plane
algebra-precalculus logarithms graphing-functions
asked Jul 27 at 15:43
roman
4341412
edited Jul 27 at 16:09
asked Jul 27 at 15:43
roman
4341412
asked Jul 27 at 15:43
roman
4341412
asked Jul 27 at 15:43
roman
4341412
4341412
Helps to note that $log_aX=log_a Yimplies X=Y$ (and conversely, if $X,Y>0$).
– lulu
Jul 27 at 15:45
Your formula involves no absolute value.
– Bernard
Jul 27 at 15:51
Keep in mind that injectivity for $log$ is essential to obtain the result.
– gimusi
Jul 27 at 16:10
add a comment |Â
Helps to note that $log_aX=log_a Yimplies X=Y$ (and conversely, if $X,Y>0$).
– lulu
Jul 27 at 15:45
Your formula involves no absolute value.
– Bernard
Jul 27 at 15:51
Keep in mind that injectivity for $log$ is essential to obtain the result.
– gimusi
Jul 27 at 16:10
Helps to note that $log_aX=log_a Yimplies X=Y$ (and conversely, if $X,Y>0$).
– lulu
Jul 27 at 15:45
Helps to note that $log_aX=log_a Yimplies X=Y$ (and conversely, if $X,Y>0$).
– lulu
Jul 27 at 15:45
Your formula involves no absolute value.
– Bernard
Jul 27 at 15:51
Your formula involves no absolute value.
– Bernard
Jul 27 at 15:51
Keep in mind that injectivity for $log$ is essential to obtain the result.
– gimusi
Jul 27 at 16:10
Keep in mind that injectivity for $log$ is essential to obtain the result.
– gimusi
Jul 27 at 16:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The given equation implies $xy-1=1-x-y+xy$ i.e. $x+y=2$. For real variables, the maximum obtainable $xy=1-(1-x)^2$ is $1$, i.e. we cannot have $xy>1$. There are therefore no real $x,,y$ that give both sides of the equation the same finite real value.
answered Jul 27 at 15:53
J.G.
13.1k11423
So the graph is just an empty plane, right? Because the only possible solution is on the "edge" of the domain.
– roman
Jul 27 at 16:02
add a comment |Â
up vote
1
down vote
Since $log x$ is injective and defined for $x>0$ we have
$$log_10(xy-1) = log_10((1-x)(1-y)) implies xy-1 = (1-x)(1-y) implies x+y-2=0$$
with the conditions
- $xy-1>0$
- $(1-x)(1-y)>0$
answered Jul 27 at 15:54
gimusi
64.9k73483
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The given equation implies $xy-1=1-x-y+xy$ i.e. $x+y=2$. For real variables, the maximum obtainable $xy=1-(1-x)^2$ is $1$, i.e. we cannot have $xy>1$. There are therefore no real $x,,y$ that give both sides of the equation the same finite real value.
answered Jul 27 at 15:53
J.G.
13.1k11423
So the graph is just an empty plane, right? Because the only possible solution is on the "edge" of the domain.
– roman
Jul 27 at 16:02
add a comment |Â
up vote
1
down vote
accepted
The given equation implies $xy-1=1-x-y+xy$ i.e. $x+y=2$. For real variables, the maximum obtainable $xy=1-(1-x)^2$ is $1$, i.e. we cannot have $xy>1$. There are therefore no real $x,,y$ that give both sides of the equation the same finite real value.
answered Jul 27 at 15:53
J.G.
13.1k11423
So the graph is just an empty plane, right? Because the only possible solution is on the "edge" of the domain.
– roman
Jul 27 at 16:02
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The given equation implies $xy-1=1-x-y+xy$ i.e. $x+y=2$. For real variables, the maximum obtainable $xy=1-(1-x)^2$ is $1$, i.e. we cannot have $xy>1$. There are therefore no real $x,,y$ that give both sides of the equation the same finite real value.
answered Jul 27 at 15:53
J.G.
13.1k11423
The given equation implies $xy-1=1-x-y+xy$ i.e. $x+y=2$. For real variables, the maximum obtainable $xy=1-(1-x)^2$ is $1$, i.e. we cannot have $xy>1$. There are therefore no real $x,,y$ that give both sides of the equation the same finite real value.
answered Jul 27 at 15:53
J.G.
13.1k11423
answered Jul 27 at 15:53
J.G.
13.1k11423
answered Jul 27 at 15:53
J.G.
13.1k11423
answered Jul 27 at 15:53
J.G.
13.1k11423
13.1k11423
So the graph is just an empty plane, right? Because the only possible solution is on the "edge" of the domain.
– roman
Jul 27 at 16:02
add a comment |Â
So the graph is just an empty plane, right? Because the only possible solution is on the "edge" of the domain.
– roman
Jul 27 at 16:02
So the graph is just an empty plane, right? Because the only possible solution is on the "edge" of the domain.
– roman
Jul 27 at 16:02
So the graph is just an empty plane, right? Because the only possible solution is on the "edge" of the domain.
– roman
Jul 27 at 16:02
add a comment |Â
up vote
1
down vote
Since $log x$ is injective and defined for $x>0$ we have
$$log_10(xy-1) = log_10((1-x)(1-y)) implies xy-1 = (1-x)(1-y) implies x+y-2=0$$
with the conditions
- $xy-1>0$
- $(1-x)(1-y)>0$
answered Jul 27 at 15:54
gimusi
64.9k73483
add a comment |Â
up vote
1
down vote
Since $log x$ is injective and defined for $x>0$ we have
$$log_10(xy-1) = log_10((1-x)(1-y)) implies xy-1 = (1-x)(1-y) implies x+y-2=0$$
with the conditions
- $xy-1>0$
- $(1-x)(1-y)>0$
answered Jul 27 at 15:54
gimusi
64.9k73483
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since $log x$ is injective and defined for $x>0$ we have
$$log_10(xy-1) = log_10((1-x)(1-y)) implies xy-1 = (1-x)(1-y) implies x+y-2=0$$
with the conditions
- $xy-1>0$
- $(1-x)(1-y)>0$
answered Jul 27 at 15:54
gimusi
64.9k73483
Since $log x$ is injective and defined for $x>0$ we have
$$log_10(xy-1) = log_10((1-x)(1-y)) implies xy-1 = (1-x)(1-y) implies x+y-2=0$$
with the conditions
- $xy-1>0$
- $(1-x)(1-y)>0$
answered Jul 27 at 15:54
gimusi
64.9k73483
answered Jul 27 at 15:54
gimusi
64.9k73483
answered Jul 27 at 15:54
gimusi
64.9k73483
answered Jul 27 at 15:54
gimusi
64.9k73483
64.9k73483
add a comment |Â
add a comment |Â
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Helps to note that $log_aX=log_a Yimplies X=Y$ (and conversely, if $X,Y>0$).
– lulu
Jul 27 at 15:45
Your formula involves no absolute value.
– Bernard
Jul 27 at 15:51
Keep in mind that injectivity for $log$ is essential to obtain the result.
– gimusi
Jul 27 at 16:10