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Evaluating $frac x y $, given $frac1x^y+1 = y$
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$$dfrac1x^y+1 = y$$
Evaluate
$$dfrac x y $$
We know that
$$dfrac1x^y+1 = x^-(y+1)$$
Which yields
$$x^-y-1 = y$$
Multiplying the both sides by $-1$
$$x^y+1 = dfrac 1 y$$
Multiplying the both sides by $x$
$$x^xy+1 = dfrac x y$$
proof-verification exponential-function
edited Jul 31 at 13:57
amWhy
189k25219431
asked Jul 31 at 13:49
Maxwell
276
add a comment |Â
up vote
-1
down vote
favorite
$$dfrac1x^y+1 = y$$
Evaluate
$$dfrac x y $$
We know that
$$dfrac1x^y+1 = x^-(y+1)$$
Which yields
$$x^-y-1 = y$$
Multiplying the both sides by $-1$
$$x^y+1 = dfrac 1 y$$
Multiplying the both sides by $x$
$$x^xy+1 = dfrac x y$$
proof-verification exponential-function
edited Jul 31 at 13:57
amWhy
189k25219431
asked Jul 31 at 13:49
Maxwell
276
Are these real numbers?
– Dr. Sonnhard Graubner
Jul 31 at 13:52
2
What should $frac x y$ be evaluated in terms of?
– Rushabh Mehta
Jul 31 at 13:52
$(x^y+1)cdot x = x^y+2$ not $x^xy+1$. What you seem to have been thinking of might have been $(x^y+1)^x$ but that would have simplified to $x^(y+1)x=x^xy+x$ not $x^xy+1$
– JMoravitz
Jul 31 at 13:55
@Dr.SonnhardGraubner They are.
– Maxwell
Jul 31 at 14:00
In any case, RushabhMehta brings up a very valid point. Normally for problems like this, we try to express the answer in terms of the fewest variables (or expressed in specific variables). $fracxy$ and $x^y+2$ are both written in terms of two variables, so neither accomplishes the goal of reducing the number of variables. Why should $x^y+2$ be a more desirable final answer than $fracxy$ itself? (I know it is silly to give $fracxy$ as an answer to the question of "express $fracxy$" but it isn't "wrong" per se, and no more or less correct than $x^y+2$ is)
– JMoravitz
Jul 31 at 14:00
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$$dfrac1x^y+1 = y$$
Evaluate
$$dfrac x y $$
We know that
$$dfrac1x^y+1 = x^-(y+1)$$
Which yields
$$x^-y-1 = y$$
Multiplying the both sides by $-1$
$$x^y+1 = dfrac 1 y$$
Multiplying the both sides by $x$
$$x^xy+1 = dfrac x y$$
proof-verification exponential-function
edited Jul 31 at 13:57
amWhy
189k25219431
asked Jul 31 at 13:49
Maxwell
276
$$dfrac1x^y+1 = y$$
Evaluate
$$dfrac x y $$
We know that
$$dfrac1x^y+1 = x^-(y+1)$$
Which yields
$$x^-y-1 = y$$
Multiplying the both sides by $-1$
$$x^y+1 = dfrac 1 y$$
Multiplying the both sides by $x$
$$x^xy+1 = dfrac x y$$
proof-verification exponential-function
edited Jul 31 at 13:57
amWhy
189k25219431
asked Jul 31 at 13:49
Maxwell
276
edited Jul 31 at 13:57
amWhy
189k25219431
edited Jul 31 at 13:57
amWhy
189k25219431
edited Jul 31 at 13:57
amWhy
189k25219431
189k25219431
asked Jul 31 at 13:49
Maxwell
276
asked Jul 31 at 13:49
Maxwell
276
asked Jul 31 at 13:49
Maxwell
276
276
Are these real numbers?
– Dr. Sonnhard Graubner
Jul 31 at 13:52
2
What should $frac x y$ be evaluated in terms of?
– Rushabh Mehta
Jul 31 at 13:52
$(x^y+1)cdot x = x^y+2$ not $x^xy+1$. What you seem to have been thinking of might have been $(x^y+1)^x$ but that would have simplified to $x^(y+1)x=x^xy+x$ not $x^xy+1$
– JMoravitz
Jul 31 at 13:55
@Dr.SonnhardGraubner They are.
– Maxwell
Jul 31 at 14:00
In any case, RushabhMehta brings up a very valid point. Normally for problems like this, we try to express the answer in terms of the fewest variables (or expressed in specific variables). $fracxy$ and $x^y+2$ are both written in terms of two variables, so neither accomplishes the goal of reducing the number of variables. Why should $x^y+2$ be a more desirable final answer than $fracxy$ itself? (I know it is silly to give $fracxy$ as an answer to the question of "express $fracxy$" but it isn't "wrong" per se, and no more or less correct than $x^y+2$ is)
– JMoravitz
Jul 31 at 14:00
add a comment |Â
Are these real numbers?
– Dr. Sonnhard Graubner
Jul 31 at 13:52
2
What should $frac x y$ be evaluated in terms of?
– Rushabh Mehta
Jul 31 at 13:52
$(x^y+1)cdot x = x^y+2$ not $x^xy+1$. What you seem to have been thinking of might have been $(x^y+1)^x$ but that would have simplified to $x^(y+1)x=x^xy+x$ not $x^xy+1$
– JMoravitz
Jul 31 at 13:55
@Dr.SonnhardGraubner They are.
– Maxwell
Jul 31 at 14:00
In any case, RushabhMehta brings up a very valid point. Normally for problems like this, we try to express the answer in terms of the fewest variables (or expressed in specific variables). $fracxy$ and $x^y+2$ are both written in terms of two variables, so neither accomplishes the goal of reducing the number of variables. Why should $x^y+2$ be a more desirable final answer than $fracxy$ itself? (I know it is silly to give $fracxy$ as an answer to the question of "express $fracxy$" but it isn't "wrong" per se, and no more or less correct than $x^y+2$ is)
– JMoravitz
Jul 31 at 14:00
Are these real numbers?
– Dr. Sonnhard Graubner
Jul 31 at 13:52
Are these real numbers?
– Dr. Sonnhard Graubner
Jul 31 at 13:52
2
2
What should $frac x y$ be evaluated in terms of?
– Rushabh Mehta
Jul 31 at 13:52
What should $frac x y$ be evaluated in terms of?
– Rushabh Mehta
Jul 31 at 13:52
$(x^y+1)cdot x = x^y+2$ not $x^xy+1$. What you seem to have been thinking of might have been $(x^y+1)^x$ but that would have simplified to $x^(y+1)x=x^xy+x$ not $x^xy+1$
– JMoravitz
Jul 31 at 13:55
$(x^y+1)cdot x = x^y+2$ not $x^xy+1$. What you seem to have been thinking of might have been $(x^y+1)^x$ but that would have simplified to $x^(y+1)x=x^xy+x$ not $x^xy+1$
– JMoravitz
Jul 31 at 13:55
@Dr.SonnhardGraubner They are.
– Maxwell
Jul 31 at 14:00
@Dr.SonnhardGraubner They are.
– Maxwell
Jul 31 at 14:00
In any case, RushabhMehta brings up a very valid point. Normally for problems like this, we try to express the answer in terms of the fewest variables (or expressed in specific variables). $fracxy$ and $x^y+2$ are both written in terms of two variables, so neither accomplishes the goal of reducing the number of variables. Why should $x^y+2$ be a more desirable final answer than $fracxy$ itself? (I know it is silly to give $fracxy$ as an answer to the question of "express $fracxy$" but it isn't "wrong" per se, and no more or less correct than $x^y+2$ is)
– JMoravitz
Jul 31 at 14:00
In any case, RushabhMehta brings up a very valid point. Normally for problems like this, we try to express the answer in terms of the fewest variables (or expressed in specific variables). $fracxy$ and $x^y+2$ are both written in terms of two variables, so neither accomplishes the goal of reducing the number of variables. Why should $x^y+2$ be a more desirable final answer than $fracxy$ itself? (I know it is silly to give $fracxy$ as an answer to the question of "express $fracxy$" but it isn't "wrong" per se, and no more or less correct than $x^y+2$ is)
– JMoravitz
Jul 31 at 14:00
add a comment |Â
1 Answer
1
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oldest
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up vote
0
down vote
Your final step is not correct. $acdot a^b=a^1cdot a^b = a^1+b$. Multiplying numbers with the same base is additive in the exponent, not multiplicative. So, $xcdot x^y+1 = x^1cdot x^y+1 = x^1+y+1 = x^y+2$.
Putting this all together:
$$dfracxy = x^y+2$$
In terms of a single variable, you have
$x=left(dfrac1yright)^tfrac1y+1$
So, that gives you:
$$dfracxy = left(dfrac1yright)^tfracy+2y+1$$
answered Jul 31 at 13:56
InterstellarProbe
2,650516
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your final step is not correct. $acdot a^b=a^1cdot a^b = a^1+b$. Multiplying numbers with the same base is additive in the exponent, not multiplicative. So, $xcdot x^y+1 = x^1cdot x^y+1 = x^1+y+1 = x^y+2$.
Putting this all together:
$$dfracxy = x^y+2$$
In terms of a single variable, you have
$x=left(dfrac1yright)^tfrac1y+1$
So, that gives you:
$$dfracxy = left(dfrac1yright)^tfracy+2y+1$$
answered Jul 31 at 13:56
InterstellarProbe
2,650516
add a comment |Â
up vote
0
down vote
Your final step is not correct. $acdot a^b=a^1cdot a^b = a^1+b$. Multiplying numbers with the same base is additive in the exponent, not multiplicative. So, $xcdot x^y+1 = x^1cdot x^y+1 = x^1+y+1 = x^y+2$.
Putting this all together:
$$dfracxy = x^y+2$$
In terms of a single variable, you have
$x=left(dfrac1yright)^tfrac1y+1$
So, that gives you:
$$dfracxy = left(dfrac1yright)^tfracy+2y+1$$
answered Jul 31 at 13:56
InterstellarProbe
2,650516
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your final step is not correct. $acdot a^b=a^1cdot a^b = a^1+b$. Multiplying numbers with the same base is additive in the exponent, not multiplicative. So, $xcdot x^y+1 = x^1cdot x^y+1 = x^1+y+1 = x^y+2$.
Putting this all together:
$$dfracxy = x^y+2$$
In terms of a single variable, you have
$x=left(dfrac1yright)^tfrac1y+1$
So, that gives you:
$$dfracxy = left(dfrac1yright)^tfracy+2y+1$$
answered Jul 31 at 13:56
InterstellarProbe
2,650516
Your final step is not correct. $acdot a^b=a^1cdot a^b = a^1+b$. Multiplying numbers with the same base is additive in the exponent, not multiplicative. So, $xcdot x^y+1 = x^1cdot x^y+1 = x^1+y+1 = x^y+2$.
Putting this all together:
$$dfracxy = x^y+2$$
In terms of a single variable, you have
$x=left(dfrac1yright)^tfrac1y+1$
So, that gives you:
$$dfracxy = left(dfrac1yright)^tfracy+2y+1$$
answered Jul 31 at 13:56
InterstellarProbe
2,650516
edited Jul 31 at 15:50
answered Jul 31 at 13:56
InterstellarProbe
2,650516
answered Jul 31 at 13:56
InterstellarProbe
2,650516
answered Jul 31 at 13:56
InterstellarProbe
2,650516
2,650516
add a comment |Â
add a comment |Â
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Are these real numbers?
– Dr. Sonnhard Graubner
Jul 31 at 13:52
2
What should $frac x y$ be evaluated in terms of?
– Rushabh Mehta
Jul 31 at 13:52
$(x^y+1)cdot x = x^y+2$ not $x^xy+1$. What you seem to have been thinking of might have been $(x^y+1)^x$ but that would have simplified to $x^(y+1)x=x^xy+x$ not $x^xy+1$
– JMoravitz
Jul 31 at 13:55
@Dr.SonnhardGraubner They are.
– Maxwell
Jul 31 at 14:00
In any case, RushabhMehta brings up a very valid point. Normally for problems like this, we try to express the answer in terms of the fewest variables (or expressed in specific variables). $fracxy$ and $x^y+2$ are both written in terms of two variables, so neither accomplishes the goal of reducing the number of variables. Why should $x^y+2$ be a more desirable final answer than $fracxy$ itself? (I know it is silly to give $fracxy$ as an answer to the question of "express $fracxy$" but it isn't "wrong" per se, and no more or less correct than $x^y+2$ is)
– JMoravitz
Jul 31 at 14:00