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For what $a$ would the graphs of $y=a^x$ and $y=log_a x$ be tangent?
Clash Royale CLAN TAG#URR8PPP
up vote
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Yesterday a question popped in my mind before going to sleep and I can't solve it (probably because my math knowledge is lacking).
Assuming there is a function
$$f(x) = a^x$$
Its inverse function would look like this:
$$f^-1(x) = log_a x$$
Question:
For which value of $a$ would the graphs of these two functions just touch at only one common point (of tangency)?
logarithms
edited Jul 24 at 15:22
Blue
43.6k868141
asked Jul 24 at 15:10
Agabagugba
384
add a comment |Â
up vote
5
down vote
favorite
Yesterday a question popped in my mind before going to sleep and I can't solve it (probably because my math knowledge is lacking).
Assuming there is a function
$$f(x) = a^x$$
Its inverse function would look like this:
$$f^-1(x) = log_a x$$
Question:
For which value of $a$ would the graphs of these two functions just touch at only one common point (of tangency)?
logarithms
edited Jul 24 at 15:22
Blue
43.6k868141
asked Jul 24 at 15:10
Agabagugba
384
2
I would prefer that you wrote $f(x)=a^x$ and $f^-1(y)=log_a y$. Also notice this forces $a>0, y>0$.
– Kevin
Jul 24 at 15:18
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Yesterday a question popped in my mind before going to sleep and I can't solve it (probably because my math knowledge is lacking).
Assuming there is a function
$$f(x) = a^x$$
Its inverse function would look like this:
$$f^-1(x) = log_a x$$
Question:
For which value of $a$ would the graphs of these two functions just touch at only one common point (of tangency)?
logarithms
edited Jul 24 at 15:22
Blue
43.6k868141
asked Jul 24 at 15:10
Agabagugba
384
Yesterday a question popped in my mind before going to sleep and I can't solve it (probably because my math knowledge is lacking).
Assuming there is a function
$$f(x) = a^x$$
Its inverse function would look like this:
$$f^-1(x) = log_a x$$
Question:
For which value of $a$ would the graphs of these two functions just touch at only one common point (of tangency)?
logarithms
edited Jul 24 at 15:22
Blue
43.6k868141
asked Jul 24 at 15:10
Agabagugba
384
edited Jul 24 at 15:22
Blue
43.6k868141
edited Jul 24 at 15:22
Blue
43.6k868141
edited Jul 24 at 15:22
Blue
43.6k868141
43.6k868141
asked Jul 24 at 15:10
Agabagugba
384
asked Jul 24 at 15:10
Agabagugba
384
asked Jul 24 at 15:10
Agabagugba
384
384
2
I would prefer that you wrote $f(x)=a^x$ and $f^-1(y)=log_a y$. Also notice this forces $a>0, y>0$.
– Kevin
Jul 24 at 15:18
add a comment |Â
2
I would prefer that you wrote $f(x)=a^x$ and $f^-1(y)=log_a y$. Also notice this forces $a>0, y>0$.
– Kevin
Jul 24 at 15:18
2
2
I would prefer that you wrote $f(x)=a^x$ and $f^-1(y)=log_a y$. Also notice this forces $a>0, y>0$.
– Kevin
Jul 24 at 15:18
I would prefer that you wrote $f(x)=a^x$ and $f^-1(y)=log_a y$. Also notice this forces $a>0, y>0$.
– Kevin
Jul 24 at 15:18
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
Such $a$ likely exists and is somewhere in the following range which can be narrowed down further: $$1.44lt a lt 1.45$$
EDIT: $$aapprox1.444667861, qquad xapprox 2.71827$$
Note that $x$ value is very close to $e$. I don't know how to explain it but it makes this problem really, really interesting!
answered Jul 24 at 15:47
Oldboy
2,6101316
Thank you very much for your answer! But is there any possible way to solve this without using a graphing program?
– Agabagugba
Jul 24 at 16:00
I doubt it, really.
– Oldboy
Jul 24 at 16:20
Oh okay.. Thank you anyways
– Agabagugba
Jul 24 at 16:33
I think the key question is: prove or disprove that touching point has $x=e$ . That would be something :) I feel like I'm missing something obvious.
– Oldboy
Jul 24 at 16:36
add a comment |Â
up vote
6
down vote
Basically, the inverse function is obtained by finding the mirror image of $y=a^x$, in $y=x $ .Now note for both the function and its inverse to intersect at just one point, $y=x $ must be the common tangent to both of them. So slope of tangent of $y=a^x $ ,$y=log_a x$ and $y=x $ must be equal. So,$ 1=a^x ln a=ln xlog_a e $ now you have 3 equations and 2 unknowns one being 'a' and other being the value of $x $ at which the functions have only one point of intersection . solve finally to get $a^e=e $ which gives the value of $a $ approximately as 1.44.
answered Jul 24 at 16:40
Jasmine
322111
Excellent, +1. I love MSE because it's constantly reminding me how dumb I know to be from time to time.
– Oldboy
Jul 24 at 18:21
add a comment |Â
up vote
2
down vote
According to a OEIS comment, the only base $a$ such that $exists xi in A subsetmathbb R: f(xi)=f^-1(xi)$ is $a=e^1/e$ with $xi=e$. Numerically this seems to be true, as Oldboy pointed out.
Whether that's unique (for any $a$) needs to be proven.
answered Jul 24 at 16:36
edmz
381211
add a comment |Â
up vote
2
down vote
As @Jasmine point out, because that $a^x$ and $log_ax$ are inverses we can add $x$ to get $x=a^x=log_ax$.
We can say more from this: $a^x=log_a ximplies a^a^x=x$ but also $x=a^x$ hence $a^a^x=log_a ximplies a^a^a^x$ and so on.
Hence we are searching for $a,x$ such that the sequence $x,a^x,a^a^x,cdots$ is a constant, $ainBbb R^+setminus1,xin Bbb R^+$.
As it is shown here we have that $e^-e leq a leq e^frac1e$ for $a^a^.^.^.$ to converge, hence, by taking the sequence and taking the $1/x$ power of it we get $x^frac1xin[e^-e,e^frac1e]$, by guessing $e$ we can find that $x=e,a=e^frac1e$ works. But $x=e$ is not the only $x$ that works. Take any $x$ such that $x^frac1xin[e^-e,e^frac1e]$ it can be shown that $x,a=x^frac1x$ are tuple that answer the condition of the post.
Edit, I realized that I made a mistake.
[$(a^a^...^x)^1/x=(a^a^...^x/x)ne (a^a^...)$, although it is interesting to see that if you continue with the logic of the mistake and find the $x$s where the 2 function "kiss" you get correct answer]
But given $x=a^x$ we have $x^1/x=a$(In analogy to the mistake, the maximum value of $x^1/x$ is $e^1/e$ at $x=e$).
As I said in the post $a=x^frac1x$ gives us $a^y=log_ay=y$, at $y=x$, but we also want them to be tangent so $(a^y)'mid_y=x=ln(a)a^x=frac1ln(a)x=(log_a(y))'mid_y=x$, we know that $a^x=x$ so $ln(a)a^x=ln(a)x$, let set $ln(a)x=z$ and we have $z=frac1zimplies z=pm1$, and so $ln(a)x=pm1impliesln(x^1/x)x=pm1impliesln(x)=pm1implies x=begincasese\1/eendcases\implies a=begincasese^1/e\(1/e)^eendcases$
(I think that OEIS made a mistake about uniqueness)
See here the graph of the four functions
answered Jul 24 at 17:09
Holo
4,1672528
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Such $a$ likely exists and is somewhere in the following range which can be narrowed down further: $$1.44lt a lt 1.45$$
EDIT: $$aapprox1.444667861, qquad xapprox 2.71827$$
Note that $x$ value is very close to $e$. I don't know how to explain it but it makes this problem really, really interesting!
answered Jul 24 at 15:47
Oldboy
2,6101316
Thank you very much for your answer! But is there any possible way to solve this without using a graphing program?
– Agabagugba
Jul 24 at 16:00
I doubt it, really.
– Oldboy
Jul 24 at 16:20
Oh okay.. Thank you anyways
– Agabagugba
Jul 24 at 16:33
I think the key question is: prove or disprove that touching point has $x=e$ . That would be something :) I feel like I'm missing something obvious.
– Oldboy
Jul 24 at 16:36
add a comment |Â
up vote
1
down vote
accepted
Such $a$ likely exists and is somewhere in the following range which can be narrowed down further: $$1.44lt a lt 1.45$$
EDIT: $$aapprox1.444667861, qquad xapprox 2.71827$$
Note that $x$ value is very close to $e$. I don't know how to explain it but it makes this problem really, really interesting!
answered Jul 24 at 15:47
Oldboy
2,6101316
Thank you very much for your answer! But is there any possible way to solve this without using a graphing program?
– Agabagugba
Jul 24 at 16:00
I doubt it, really.
– Oldboy
Jul 24 at 16:20
Oh okay.. Thank you anyways
– Agabagugba
Jul 24 at 16:33
I think the key question is: prove or disprove that touching point has $x=e$ . That would be something :) I feel like I'm missing something obvious.
– Oldboy
Jul 24 at 16:36
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Such $a$ likely exists and is somewhere in the following range which can be narrowed down further: $$1.44lt a lt 1.45$$
EDIT: $$aapprox1.444667861, qquad xapprox 2.71827$$
Note that $x$ value is very close to $e$. I don't know how to explain it but it makes this problem really, really interesting!
answered Jul 24 at 15:47
Oldboy
2,6101316
Such $a$ likely exists and is somewhere in the following range which can be narrowed down further: $$1.44lt a lt 1.45$$
EDIT: $$aapprox1.444667861, qquad xapprox 2.71827$$
Note that $x$ value is very close to $e$. I don't know how to explain it but it makes this problem really, really interesting!
answered Jul 24 at 15:47
Oldboy
2,6101316
edited Jul 24 at 16:34
answered Jul 24 at 15:47
Oldboy
2,6101316
answered Jul 24 at 15:47
Oldboy
2,6101316
answered Jul 24 at 15:47
Oldboy
2,6101316
2,6101316
Thank you very much for your answer! But is there any possible way to solve this without using a graphing program?
– Agabagugba
Jul 24 at 16:00
I doubt it, really.
– Oldboy
Jul 24 at 16:20
Oh okay.. Thank you anyways
– Agabagugba
Jul 24 at 16:33
I think the key question is: prove or disprove that touching point has $x=e$ . That would be something :) I feel like I'm missing something obvious.
– Oldboy
Jul 24 at 16:36
add a comment |Â
Thank you very much for your answer! But is there any possible way to solve this without using a graphing program?
– Agabagugba
Jul 24 at 16:00
I doubt it, really.
– Oldboy
Jul 24 at 16:20
Oh okay.. Thank you anyways
– Agabagugba
Jul 24 at 16:33
I think the key question is: prove or disprove that touching point has $x=e$ . That would be something :) I feel like I'm missing something obvious.
– Oldboy
Jul 24 at 16:36
Thank you very much for your answer! But is there any possible way to solve this without using a graphing program?
– Agabagugba
Jul 24 at 16:00
Thank you very much for your answer! But is there any possible way to solve this without using a graphing program?
– Agabagugba
Jul 24 at 16:00
I doubt it, really.
– Oldboy
Jul 24 at 16:20
I doubt it, really.
– Oldboy
Jul 24 at 16:20
Oh okay.. Thank you anyways
– Agabagugba
Jul 24 at 16:33
Oh okay.. Thank you anyways
– Agabagugba
Jul 24 at 16:33
I think the key question is: prove or disprove that touching point has $x=e$ . That would be something :) I feel like I'm missing something obvious.
– Oldboy
Jul 24 at 16:36
I think the key question is: prove or disprove that touching point has $x=e$ . That would be something :) I feel like I'm missing something obvious.
– Oldboy
Jul 24 at 16:36
add a comment |Â
up vote
6
down vote
Basically, the inverse function is obtained by finding the mirror image of $y=a^x$, in $y=x $ .Now note for both the function and its inverse to intersect at just one point, $y=x $ must be the common tangent to both of them. So slope of tangent of $y=a^x $ ,$y=log_a x$ and $y=x $ must be equal. So,$ 1=a^x ln a=ln xlog_a e $ now you have 3 equations and 2 unknowns one being 'a' and other being the value of $x $ at which the functions have only one point of intersection . solve finally to get $a^e=e $ which gives the value of $a $ approximately as 1.44.
answered Jul 24 at 16:40
Jasmine
322111
Excellent, +1. I love MSE because it's constantly reminding me how dumb I know to be from time to time.
– Oldboy
Jul 24 at 18:21
add a comment |Â
up vote
6
down vote
Basically, the inverse function is obtained by finding the mirror image of $y=a^x$, in $y=x $ .Now note for both the function and its inverse to intersect at just one point, $y=x $ must be the common tangent to both of them. So slope of tangent of $y=a^x $ ,$y=log_a x$ and $y=x $ must be equal. So,$ 1=a^x ln a=ln xlog_a e $ now you have 3 equations and 2 unknowns one being 'a' and other being the value of $x $ at which the functions have only one point of intersection . solve finally to get $a^e=e $ which gives the value of $a $ approximately as 1.44.
answered Jul 24 at 16:40
Jasmine
322111
Excellent, +1. I love MSE because it's constantly reminding me how dumb I know to be from time to time.
– Oldboy
Jul 24 at 18:21
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Basically, the inverse function is obtained by finding the mirror image of $y=a^x$, in $y=x $ .Now note for both the function and its inverse to intersect at just one point, $y=x $ must be the common tangent to both of them. So slope of tangent of $y=a^x $ ,$y=log_a x$ and $y=x $ must be equal. So,$ 1=a^x ln a=ln xlog_a e $ now you have 3 equations and 2 unknowns one being 'a' and other being the value of $x $ at which the functions have only one point of intersection . solve finally to get $a^e=e $ which gives the value of $a $ approximately as 1.44.
answered Jul 24 at 16:40
Jasmine
322111
Basically, the inverse function is obtained by finding the mirror image of $y=a^x$, in $y=x $ .Now note for both the function and its inverse to intersect at just one point, $y=x $ must be the common tangent to both of them. So slope of tangent of $y=a^x $ ,$y=log_a x$ and $y=x $ must be equal. So,$ 1=a^x ln a=ln xlog_a e $ now you have 3 equations and 2 unknowns one being 'a' and other being the value of $x $ at which the functions have only one point of intersection . solve finally to get $a^e=e $ which gives the value of $a $ approximately as 1.44.
answered Jul 24 at 16:40
Jasmine
322111
edited Jul 24 at 17:00
answered Jul 24 at 16:40
Jasmine
322111
answered Jul 24 at 16:40
Jasmine
322111
answered Jul 24 at 16:40
Jasmine
322111
322111
Excellent, +1. I love MSE because it's constantly reminding me how dumb I know to be from time to time.
– Oldboy
Jul 24 at 18:21
add a comment |Â
Excellent, +1. I love MSE because it's constantly reminding me how dumb I know to be from time to time.
– Oldboy
Jul 24 at 18:21
Excellent, +1. I love MSE because it's constantly reminding me how dumb I know to be from time to time.
– Oldboy
Jul 24 at 18:21
Excellent, +1. I love MSE because it's constantly reminding me how dumb I know to be from time to time.
– Oldboy
Jul 24 at 18:21
add a comment |Â
up vote
2
down vote
According to a OEIS comment, the only base $a$ such that $exists xi in A subsetmathbb R: f(xi)=f^-1(xi)$ is $a=e^1/e$ with $xi=e$. Numerically this seems to be true, as Oldboy pointed out.
Whether that's unique (for any $a$) needs to be proven.
answered Jul 24 at 16:36
edmz
381211
add a comment |Â
up vote
2
down vote
According to a OEIS comment, the only base $a$ such that $exists xi in A subsetmathbb R: f(xi)=f^-1(xi)$ is $a=e^1/e$ with $xi=e$. Numerically this seems to be true, as Oldboy pointed out.
Whether that's unique (for any $a$) needs to be proven.
answered Jul 24 at 16:36
edmz
381211
add a comment |Â
up vote
2
down vote
up vote
2
down vote
According to a OEIS comment, the only base $a$ such that $exists xi in A subsetmathbb R: f(xi)=f^-1(xi)$ is $a=e^1/e$ with $xi=e$. Numerically this seems to be true, as Oldboy pointed out.
Whether that's unique (for any $a$) needs to be proven.
answered Jul 24 at 16:36
edmz
381211
According to a OEIS comment, the only base $a$ such that $exists xi in A subsetmathbb R: f(xi)=f^-1(xi)$ is $a=e^1/e$ with $xi=e$. Numerically this seems to be true, as Oldboy pointed out.
Whether that's unique (for any $a$) needs to be proven.
answered Jul 24 at 16:36
edmz
381211
answered Jul 24 at 16:36
edmz
381211
answered Jul 24 at 16:36
edmz
381211
answered Jul 24 at 16:36
edmz
381211
381211
add a comment |Â
add a comment |Â
up vote
2
down vote
As @Jasmine point out, because that $a^x$ and $log_ax$ are inverses we can add $x$ to get $x=a^x=log_ax$.
We can say more from this: $a^x=log_a ximplies a^a^x=x$ but also $x=a^x$ hence $a^a^x=log_a ximplies a^a^a^x$ and so on.
Hence we are searching for $a,x$ such that the sequence $x,a^x,a^a^x,cdots$ is a constant, $ainBbb R^+setminus1,xin Bbb R^+$.
As it is shown here we have that $e^-e leq a leq e^frac1e$ for $a^a^.^.^.$ to converge, hence, by taking the sequence and taking the $1/x$ power of it we get $x^frac1xin[e^-e,e^frac1e]$, by guessing $e$ we can find that $x=e,a=e^frac1e$ works. But $x=e$ is not the only $x$ that works. Take any $x$ such that $x^frac1xin[e^-e,e^frac1e]$ it can be shown that $x,a=x^frac1x$ are tuple that answer the condition of the post.
Edit, I realized that I made a mistake.
[$(a^a^...^x)^1/x=(a^a^...^x/x)ne (a^a^...)$, although it is interesting to see that if you continue with the logic of the mistake and find the $x$s where the 2 function "kiss" you get correct answer]
But given $x=a^x$ we have $x^1/x=a$(In analogy to the mistake, the maximum value of $x^1/x$ is $e^1/e$ at $x=e$).
As I said in the post $a=x^frac1x$ gives us $a^y=log_ay=y$, at $y=x$, but we also want them to be tangent so $(a^y)'mid_y=x=ln(a)a^x=frac1ln(a)x=(log_a(y))'mid_y=x$, we know that $a^x=x$ so $ln(a)a^x=ln(a)x$, let set $ln(a)x=z$ and we have $z=frac1zimplies z=pm1$, and so $ln(a)x=pm1impliesln(x^1/x)x=pm1impliesln(x)=pm1implies x=begincasese\1/eendcases\implies a=begincasese^1/e\(1/e)^eendcases$
(I think that OEIS made a mistake about uniqueness)
See here the graph of the four functions
answered Jul 24 at 17:09
Holo
4,1672528
add a comment |Â
up vote
2
down vote
As @Jasmine point out, because that $a^x$ and $log_ax$ are inverses we can add $x$ to get $x=a^x=log_ax$.
We can say more from this: $a^x=log_a ximplies a^a^x=x$ but also $x=a^x$ hence $a^a^x=log_a ximplies a^a^a^x$ and so on.
Hence we are searching for $a,x$ such that the sequence $x,a^x,a^a^x,cdots$ is a constant, $ainBbb R^+setminus1,xin Bbb R^+$.
As it is shown here we have that $e^-e leq a leq e^frac1e$ for $a^a^.^.^.$ to converge, hence, by taking the sequence and taking the $1/x$ power of it we get $x^frac1xin[e^-e,e^frac1e]$, by guessing $e$ we can find that $x=e,a=e^frac1e$ works. But $x=e$ is not the only $x$ that works. Take any $x$ such that $x^frac1xin[e^-e,e^frac1e]$ it can be shown that $x,a=x^frac1x$ are tuple that answer the condition of the post.
Edit, I realized that I made a mistake.
[$(a^a^...^x)^1/x=(a^a^...^x/x)ne (a^a^...)$, although it is interesting to see that if you continue with the logic of the mistake and find the $x$s where the 2 function "kiss" you get correct answer]
But given $x=a^x$ we have $x^1/x=a$(In analogy to the mistake, the maximum value of $x^1/x$ is $e^1/e$ at $x=e$).
As I said in the post $a=x^frac1x$ gives us $a^y=log_ay=y$, at $y=x$, but we also want them to be tangent so $(a^y)'mid_y=x=ln(a)a^x=frac1ln(a)x=(log_a(y))'mid_y=x$, we know that $a^x=x$ so $ln(a)a^x=ln(a)x$, let set $ln(a)x=z$ and we have $z=frac1zimplies z=pm1$, and so $ln(a)x=pm1impliesln(x^1/x)x=pm1impliesln(x)=pm1implies x=begincasese\1/eendcases\implies a=begincasese^1/e\(1/e)^eendcases$
(I think that OEIS made a mistake about uniqueness)
See here the graph of the four functions
answered Jul 24 at 17:09
Holo
4,1672528
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As @Jasmine point out, because that $a^x$ and $log_ax$ are inverses we can add $x$ to get $x=a^x=log_ax$.
We can say more from this: $a^x=log_a ximplies a^a^x=x$ but also $x=a^x$ hence $a^a^x=log_a ximplies a^a^a^x$ and so on.
Hence we are searching for $a,x$ such that the sequence $x,a^x,a^a^x,cdots$ is a constant, $ainBbb R^+setminus1,xin Bbb R^+$.
As it is shown here we have that $e^-e leq a leq e^frac1e$ for $a^a^.^.^.$ to converge, hence, by taking the sequence and taking the $1/x$ power of it we get $x^frac1xin[e^-e,e^frac1e]$, by guessing $e$ we can find that $x=e,a=e^frac1e$ works. But $x=e$ is not the only $x$ that works. Take any $x$ such that $x^frac1xin[e^-e,e^frac1e]$ it can be shown that $x,a=x^frac1x$ are tuple that answer the condition of the post.
Edit, I realized that I made a mistake.
[$(a^a^...^x)^1/x=(a^a^...^x/x)ne (a^a^...)$, although it is interesting to see that if you continue with the logic of the mistake and find the $x$s where the 2 function "kiss" you get correct answer]
But given $x=a^x$ we have $x^1/x=a$(In analogy to the mistake, the maximum value of $x^1/x$ is $e^1/e$ at $x=e$).
As I said in the post $a=x^frac1x$ gives us $a^y=log_ay=y$, at $y=x$, but we also want them to be tangent so $(a^y)'mid_y=x=ln(a)a^x=frac1ln(a)x=(log_a(y))'mid_y=x$, we know that $a^x=x$ so $ln(a)a^x=ln(a)x$, let set $ln(a)x=z$ and we have $z=frac1zimplies z=pm1$, and so $ln(a)x=pm1impliesln(x^1/x)x=pm1impliesln(x)=pm1implies x=begincasese\1/eendcases\implies a=begincasese^1/e\(1/e)^eendcases$
(I think that OEIS made a mistake about uniqueness)
See here the graph of the four functions
answered Jul 24 at 17:09
Holo
4,1672528
As @Jasmine point out, because that $a^x$ and $log_ax$ are inverses we can add $x$ to get $x=a^x=log_ax$.
We can say more from this: $a^x=log_a ximplies a^a^x=x$ but also $x=a^x$ hence $a^a^x=log_a ximplies a^a^a^x$ and so on.
Hence we are searching for $a,x$ such that the sequence $x,a^x,a^a^x,cdots$ is a constant, $ainBbb R^+setminus1,xin Bbb R^+$.
As it is shown here we have that $e^-e leq a leq e^frac1e$ for $a^a^.^.^.$ to converge, hence, by taking the sequence and taking the $1/x$ power of it we get $x^frac1xin[e^-e,e^frac1e]$, by guessing $e$ we can find that $x=e,a=e^frac1e$ works. But $x=e$ is not the only $x$ that works. Take any $x$ such that $x^frac1xin[e^-e,e^frac1e]$ it can be shown that $x,a=x^frac1x$ are tuple that answer the condition of the post.
Edit, I realized that I made a mistake.
[$(a^a^...^x)^1/x=(a^a^...^x/x)ne (a^a^...)$, although it is interesting to see that if you continue with the logic of the mistake and find the $x$s where the 2 function "kiss" you get correct answer]
But given $x=a^x$ we have $x^1/x=a$(In analogy to the mistake, the maximum value of $x^1/x$ is $e^1/e$ at $x=e$).
As I said in the post $a=x^frac1x$ gives us $a^y=log_ay=y$, at $y=x$, but we also want them to be tangent so $(a^y)'mid_y=x=ln(a)a^x=frac1ln(a)x=(log_a(y))'mid_y=x$, we know that $a^x=x$ so $ln(a)a^x=ln(a)x$, let set $ln(a)x=z$ and we have $z=frac1zimplies z=pm1$, and so $ln(a)x=pm1impliesln(x^1/x)x=pm1impliesln(x)=pm1implies x=begincasese\1/eendcases\implies a=begincasese^1/e\(1/e)^eendcases$
(I think that OEIS made a mistake about uniqueness)
See here the graph of the four functions
answered Jul 24 at 17:09
Holo
4,1672528
edited Aug 6 at 21:46
answered Jul 24 at 17:09
Holo
4,1672528
answered Jul 24 at 17:09
Holo
4,1672528
answered Jul 24 at 17:09
Holo
4,1672528
4,1672528
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2
I would prefer that you wrote $f(x)=a^x$ and $f^-1(y)=log_a y$. Also notice this forces $a>0, y>0$.
– Kevin
Jul 24 at 15:18