Mixing
Find $y = dots$ in $x = fraclny+1lny-1$
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The problem asks to get y out of x. Like this:
$$x=logyimplies y=e^x$$
So I have this:
$$x = fraclny+1lny-1$$
This is what I've attempted so far:
$$x = fraclny+1lny-1 implies (lny-1)x = lny+1 implies xlny-x=lny+1 implies dots$$
I don't really know how to do this. I think I need to use the definition of logarithm somehow. Any hints? The answer in my textbooks is
$$y = e^fracx+1x-1$$
algebra-precalculus logarithms
edited Jul 28 at 10:29
José Carlos Santos
112k1696173
asked Jul 28 at 10:25
Cesare
35219
add a comment |Â
up vote
1
down vote
favorite
The problem asks to get y out of x. Like this:
$$x=logyimplies y=e^x$$
So I have this:
$$x = fraclny+1lny-1$$
This is what I've attempted so far:
$$x = fraclny+1lny-1 implies (lny-1)x = lny+1 implies xlny-x=lny+1 implies dots$$
I don't really know how to do this. I think I need to use the definition of logarithm somehow. Any hints? The answer in my textbooks is
$$y = e^fracx+1x-1$$
algebra-precalculus logarithms
edited Jul 28 at 10:29
José Carlos Santos
112k1696173
asked Jul 28 at 10:25
Cesare
35219
3
Two great answers already, but you'd be surprised how many problems there are like this where it helps to notice $(x+1)/(x-1)$ is its own inverse. The easiest proof is applying the function twice.
– J.G.
Jul 28 at 10:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The problem asks to get y out of x. Like this:
$$x=logyimplies y=e^x$$
So I have this:
$$x = fraclny+1lny-1$$
This is what I've attempted so far:
$$x = fraclny+1lny-1 implies (lny-1)x = lny+1 implies xlny-x=lny+1 implies dots$$
I don't really know how to do this. I think I need to use the definition of logarithm somehow. Any hints? The answer in my textbooks is
$$y = e^fracx+1x-1$$
algebra-precalculus logarithms
edited Jul 28 at 10:29
José Carlos Santos
112k1696173
asked Jul 28 at 10:25
Cesare
35219
The problem asks to get y out of x. Like this:
$$x=logyimplies y=e^x$$
So I have this:
$$x = fraclny+1lny-1$$
This is what I've attempted so far:
$$x = fraclny+1lny-1 implies (lny-1)x = lny+1 implies xlny-x=lny+1 implies dots$$
I don't really know how to do this. I think I need to use the definition of logarithm somehow. Any hints? The answer in my textbooks is
$$y = e^fracx+1x-1$$
algebra-precalculus logarithms
edited Jul 28 at 10:29
José Carlos Santos
112k1696173
asked Jul 28 at 10:25
Cesare
35219
edited Jul 28 at 10:29
José Carlos Santos
112k1696173
edited Jul 28 at 10:29
José Carlos Santos
112k1696173
edited Jul 28 at 10:29
José Carlos Santos
112k1696173
112k1696173
asked Jul 28 at 10:25
Cesare
35219
asked Jul 28 at 10:25
Cesare
35219
asked Jul 28 at 10:25
Cesare
35219
35219
3
Two great answers already, but you'd be surprised how many problems there are like this where it helps to notice $(x+1)/(x-1)$ is its own inverse. The easiest proof is applying the function twice.
– J.G.
Jul 28 at 10:37
add a comment |Â
3
Two great answers already, but you'd be surprised how many problems there are like this where it helps to notice $(x+1)/(x-1)$ is its own inverse. The easiest proof is applying the function twice.
– J.G.
Jul 28 at 10:37
3
3
Two great answers already, but you'd be surprised how many problems there are like this where it helps to notice $(x+1)/(x-1)$ is its own inverse. The easiest proof is applying the function twice.
– J.G.
Jul 28 at 10:37
Two great answers already, but you'd be surprised how many problems there are like this where it helps to notice $(x+1)/(x-1)$ is its own inverse. The easiest proof is applying the function twice.
– J.G.
Jul 28 at 10:37
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
What you've done is fine. Now, you can deduce that$$ln y=fracx+1x-1$$and that therefore$$y=expleft(fracx+1x-1right).$$
answered Jul 28 at 10:28
José Carlos Santos
112k1696173
add a comment |Â
up vote
2
down vote
Try to put the $y$ together. Following what you already did, we get (for $xnot = 1$, which is always the case given the starting equation):
$$xlny-x=lny+1 implies (x-1)operatornamelny=x+1implies operatornamelny= fracx+1x-1 implies y=e^fracx+1x-1$$
answered Jul 28 at 10:28
Suzet
2,203427
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
What you've done is fine. Now, you can deduce that$$ln y=fracx+1x-1$$and that therefore$$y=expleft(fracx+1x-1right).$$
answered Jul 28 at 10:28
José Carlos Santos
112k1696173
add a comment |Â
up vote
2
down vote
accepted
What you've done is fine. Now, you can deduce that$$ln y=fracx+1x-1$$and that therefore$$y=expleft(fracx+1x-1right).$$
answered Jul 28 at 10:28
José Carlos Santos
112k1696173
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
What you've done is fine. Now, you can deduce that$$ln y=fracx+1x-1$$and that therefore$$y=expleft(fracx+1x-1right).$$
answered Jul 28 at 10:28
José Carlos Santos
112k1696173
What you've done is fine. Now, you can deduce that$$ln y=fracx+1x-1$$and that therefore$$y=expleft(fracx+1x-1right).$$
answered Jul 28 at 10:28
José Carlos Santos
112k1696173
answered Jul 28 at 10:28
José Carlos Santos
112k1696173
answered Jul 28 at 10:28
José Carlos Santos
112k1696173
answered Jul 28 at 10:28
José Carlos Santos
112k1696173
112k1696173
add a comment |Â
add a comment |Â
up vote
2
down vote
Try to put the $y$ together. Following what you already did, we get (for $xnot = 1$, which is always the case given the starting equation):
$$xlny-x=lny+1 implies (x-1)operatornamelny=x+1implies operatornamelny= fracx+1x-1 implies y=e^fracx+1x-1$$
answered Jul 28 at 10:28
Suzet
2,203427
add a comment |Â
up vote
2
down vote
Try to put the $y$ together. Following what you already did, we get (for $xnot = 1$, which is always the case given the starting equation):
$$xlny-x=lny+1 implies (x-1)operatornamelny=x+1implies operatornamelny= fracx+1x-1 implies y=e^fracx+1x-1$$
answered Jul 28 at 10:28
Suzet
2,203427
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Try to put the $y$ together. Following what you already did, we get (for $xnot = 1$, which is always the case given the starting equation):
$$xlny-x=lny+1 implies (x-1)operatornamelny=x+1implies operatornamelny= fracx+1x-1 implies y=e^fracx+1x-1$$
answered Jul 28 at 10:28
Suzet
2,203427
Try to put the $y$ together. Following what you already did, we get (for $xnot = 1$, which is always the case given the starting equation):
$$xlny-x=lny+1 implies (x-1)operatornamelny=x+1implies operatornamelny= fracx+1x-1 implies y=e^fracx+1x-1$$
answered Jul 28 at 10:28
Suzet
2,203427
answered Jul 28 at 10:28
Suzet
2,203427
answered Jul 28 at 10:28
Suzet
2,203427
answered Jul 28 at 10:28
Suzet
2,203427
2,203427
add a comment |Â
add a comment |Â
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3
Two great answers already, but you'd be surprised how many problems there are like this where it helps to notice $(x+1)/(x-1)$ is its own inverse. The easiest proof is applying the function twice.
– J.G.
Jul 28 at 10:37