Mixing
$ln(x)=ax^4$. Find a value for $a$ such that the function has only one real root. [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
-1
down vote
favorite
Question:
$ln(x)=ax^4$. Find a value for $a$ such that the function has only one real
root.
How do i go about this kind of exercises? It is for a multiple choice exam for college admission so i only need the reason behind it in order to be able to solve such exercises. I tried plotting the graphs but I cant tell.
calculus algebra-precalculus
edited Jul 14 at 18:00
Greg Martin
34k23060
asked Jul 14 at 17:11
Radu Gabriel
1
closed as off-topic by amWhy, Mostafa Ayaz, Cameron Williams, José Carlos Santos, mechanodroid Jul 14 at 23:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Mostafa Ayaz, Cameron Williams, Jos̩ Carlos Santos, mechanodroid
add a comment |Â
up vote
-1
down vote
favorite
Question:
$ln(x)=ax^4$. Find a value for $a$ such that the function has only one real
root.
How do i go about this kind of exercises? It is for a multiple choice exam for college admission so i only need the reason behind it in order to be able to solve such exercises. I tried plotting the graphs but I cant tell.
calculus algebra-precalculus
edited Jul 14 at 18:00
Greg Martin
34k23060
asked Jul 14 at 17:11
Radu Gabriel
1
closed as off-topic by amWhy, Mostafa Ayaz, Cameron Williams, José Carlos Santos, mechanodroid Jul 14 at 23:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Mostafa Ayaz, Cameron Williams, Jos̩ Carlos Santos, mechanodroid
Are we assuming $a > 0$. Other wise $a = 0$ will yield $ln x = 0$ has only one solution.
– fleablood
Jul 14 at 18:10
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Question:
$ln(x)=ax^4$. Find a value for $a$ such that the function has only one real
root.
How do i go about this kind of exercises? It is for a multiple choice exam for college admission so i only need the reason behind it in order to be able to solve such exercises. I tried plotting the graphs but I cant tell.
calculus algebra-precalculus
edited Jul 14 at 18:00
Greg Martin
34k23060
asked Jul 14 at 17:11
Radu Gabriel
1
Question:
$ln(x)=ax^4$. Find a value for $a$ such that the function has only one real
root.
How do i go about this kind of exercises? It is for a multiple choice exam for college admission so i only need the reason behind it in order to be able to solve such exercises. I tried plotting the graphs but I cant tell.
calculus algebra-precalculus
edited Jul 14 at 18:00
Greg Martin
34k23060
asked Jul 14 at 17:11
Radu Gabriel
1
edited Jul 14 at 18:00
Greg Martin
34k23060
edited Jul 14 at 18:00
Greg Martin
34k23060
edited Jul 14 at 18:00
Greg Martin
34k23060
34k23060
asked Jul 14 at 17:11
Radu Gabriel
1
asked Jul 14 at 17:11
Radu Gabriel
1
asked Jul 14 at 17:11
Radu Gabriel
1
1
closed as off-topic by amWhy, Mostafa Ayaz, Cameron Williams, José Carlos Santos, mechanodroid Jul 14 at 23:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Mostafa Ayaz, Cameron Williams, Jos̩ Carlos Santos, mechanodroid
closed as off-topic by amWhy, Mostafa Ayaz, Cameron Williams, José Carlos Santos, mechanodroid Jul 14 at 23:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Mostafa Ayaz, Cameron Williams, Jos̩ Carlos Santos, mechanodroid
Are we assuming $a > 0$. Other wise $a = 0$ will yield $ln x = 0$ has only one solution.
– fleablood
Jul 14 at 18:10
add a comment |Â
Are we assuming $a > 0$. Other wise $a = 0$ will yield $ln x = 0$ has only one solution.
– fleablood
Jul 14 at 18:10
Are we assuming $a > 0$. Other wise $a = 0$ will yield $ln x = 0$ has only one solution.
– fleablood
Jul 14 at 18:10
Are we assuming $a > 0$. Other wise $a = 0$ will yield $ln x = 0$ has only one solution.
– fleablood
Jul 14 at 18:10
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
3
down vote
Look at the figure below
the curvature of $ax^4$ is always positive and that of $ln x$ is always negative (by twice differentiating them) therefore they intersect in exactly one point if they are also tangent in that point i.e.$$ln x=ax^4\dfrac1x=4ax^3$$which leads to $$ax^4=1over 4$$ and by substituting we have $$ln x=dfrac14$$which yields to $$x=e^1over 4$$and $$a=dfrac14e$$also the case $ale 0$ is trivial since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
answered Jul 14 at 18:02
Mostafa Ayaz
8,6723630
What if $a le 0$?
– fleablood
Jul 14 at 18:11
This is the trivial case since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
– Mostafa Ayaz
Jul 14 at 18:15
add a comment |Â
up vote
1
down vote
Solution
Obviously, $x>0$. $ln x=ax^4$ could be rewritten as $$a=fracln xx^4.$$ We want to find the range of $a$ such that $y=a$ and $y=f(x)=dfracln xx^4$ intersect at only one point.
Notice that $$f'(x)=frac1-4ln xx^5.$$ Let $y'=0$. We have $x=e^frac14.$ Morover, $f'(x)>0$ when $x<e^frac14$ and $f'(x)<0$ when $x>e^frac14.$ Thus, we may claim that $f(x)$ increases over $(0,e^frac14]$ and decrease over $[e^frac14,+infty].$ Futher, $f(x) in (-infty,dfrac14e]$ and $f(x) in (dfrac14e,0)$. Here, you should notice $$limlimits_x to +inftyfracln xx^4=limlimits_x to +inftyfrac1/ x4x^3=limlimits_x to +inftyfrac14x^4=0.$$
Now, we may obtain that the range of $a$ we want is $(-infty,0]bigcapdfrac14e$. Otherwise, there exist either two or none intersetion.
answered Jul 14 at 18:01
mengdie1982
2,972216
Thank you very much. I actually understand the reason behind such exercises but I could not figure how to start it. I got confused because a was not separated from x and could not understand why. Thank you very much, again, and everyone else who has commented.
– Radu Gabriel
Jul 14 at 18:36
Not at all. please correct me if I'm wrong.
– mengdie1982
Jul 14 at 18:45
add a comment |Â
up vote
0
down vote
It is easy to show that the function $ln x-ax^4$ has at most a single extremum, at $x=dfrac1sqrt[4]4a$.
For any $ale0$, there is no extremum and a single root, as the function goes from $-infty$ to $infty$.
If $a>0$, the value at the extremum is
$$frac14ln4a-frac14$$ and is zero for $4a=e$, forming a double root.
answered Jul 14 at 17:29
Yves Daoust
111k665205
add a comment |Â
up vote
0
down vote
Assume $a >0$.
Considering the U/valley shape of the graph of $ax^4$ and the slouching slug shape of $ln x$ it's clear that the graphs either never intersect and $ln x < ax^4$ for all $x> 0$. Or they touch once and there is one solution to $ln x = ax^4$ or the cross and recross and there are two solutions.
Now to solve $ln x = ax^4$ is to solve $h(x) = ax^4 - ln x = 0$ and as $a^4 ge ln x$ (the graphs touch once and never cross) then $h(x) = 0$ is local minimum.
So $h(x) = ax^4 - ln x = 0$ and $h'(x) = 4ax^3 - frac 1x =0$. Then later becomes $4ax^4 - 1 = 0$ and notice both $ax^4 - ln x = 0$ and $4ax^4 - 1=0$ have terms pertaining to $ax^4$. And we can solve $ax^4 = frac 14$.
So $frac 14 - ln x = 0$ and $ln x = frac 14$ and $x = e^frac 14$.
Plugging that in we get $a*e - frac 14 = 0$.
Which gives us $a = frac 14e$
If $a =0$ then $ln x = ax^4 = 0$ and $x = 0$ is the onlys solution.
If $a < 0$. Well, look at the graphs. The U valley is upside down and every value intersect the slouching slug exactly once. So $a in (-infty, 0]cup frac 14e$.
answered Jul 14 at 17:55
fleablood
60.5k22575
add a comment |Â
up vote
0
down vote
Hint: Consider the function
$f(x)=fracln(x)x^4$ and use calculus.
Use that
$$f'(x)=frac1-4ln(x)x^4$$
answered Jul 14 at 17:13
Dr. Sonnhard Graubner
67k32660
1
don't you mean $f'(x)=frac1-4ln(x)x^5$
– Bruce
Jul 14 at 17:17
Thank you for your hint!
– Dr. Sonnhard Graubner
Jul 14 at 18:21
add a comment |Â
up vote
0
down vote
beginalign
ln x&=ax^4
tag1label1
,\
-4ln x&=-4ax^4
,\
ln(x^-4)&=-4ax^4
,\
x^-4ln(x^-4)&=-4a
.
endalign
Let $ln(x^-4)=u$, then
$x^-4=exp(u)$
and we arrive at an equation
beginalign
uexp(u)&=-4a
,
endalign
which has a well-known
closed-form solution in terms of
Lambert W function.
beginalign
operatornameW(uexp(u))&=operatornameW(-4a)
,\
u&=operatornameW(-4a)
,\
ln(x^-4)&=operatornameW(-4a)
,\
x^-4&=exp(operatornameW(-4a))
,\
x&=exp(-tfrac14cdotoperatornameW(-4a))
tag2label2
.
endalign
The number of real solutions to eqref1
is defined by the number of
real solutions to eqref2,
which in turn, is defined by the argument of $operatornameW$,
and it is well-known
that we have a single real solution
when the argument of $operatornameW$
is either non-negative, or is equal to
$-tfrac1mathrme$.
That is,
beginalign
1)& -4age0quadtoquad ale0
;\
2)& -4a=-tfrac1mathrme
quadtoquad
a=tfrac1mathrm4e
approx 0.09196986
.
endalign
In the last case we have
beginalign
x&=exp(-tfrac14cdotoperatornameW(-tfrac1mathrme))
=exp(-tfrac14cdot(-1))
=sqrt[4]mathrme
.
endalign
answered Jul 14 at 19:33
g.kov
5,5521717
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Look at the figure below
the curvature of $ax^4$ is always positive and that of $ln x$ is always negative (by twice differentiating them) therefore they intersect in exactly one point if they are also tangent in that point i.e.$$ln x=ax^4\dfrac1x=4ax^3$$which leads to $$ax^4=1over 4$$ and by substituting we have $$ln x=dfrac14$$which yields to $$x=e^1over 4$$and $$a=dfrac14e$$also the case $ale 0$ is trivial since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
answered Jul 14 at 18:02
Mostafa Ayaz
8,6723630
What if $a le 0$?
– fleablood
Jul 14 at 18:11
This is the trivial case since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
– Mostafa Ayaz
Jul 14 at 18:15
add a comment |Â
up vote
3
down vote
Look at the figure below
the curvature of $ax^4$ is always positive and that of $ln x$ is always negative (by twice differentiating them) therefore they intersect in exactly one point if they are also tangent in that point i.e.$$ln x=ax^4\dfrac1x=4ax^3$$which leads to $$ax^4=1over 4$$ and by substituting we have $$ln x=dfrac14$$which yields to $$x=e^1over 4$$and $$a=dfrac14e$$also the case $ale 0$ is trivial since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
answered Jul 14 at 18:02
Mostafa Ayaz
8,6723630
What if $a le 0$?
– fleablood
Jul 14 at 18:11
This is the trivial case since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
– Mostafa Ayaz
Jul 14 at 18:15
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Look at the figure below
the curvature of $ax^4$ is always positive and that of $ln x$ is always negative (by twice differentiating them) therefore they intersect in exactly one point if they are also tangent in that point i.e.$$ln x=ax^4\dfrac1x=4ax^3$$which leads to $$ax^4=1over 4$$ and by substituting we have $$ln x=dfrac14$$which yields to $$x=e^1over 4$$and $$a=dfrac14e$$also the case $ale 0$ is trivial since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
answered Jul 14 at 18:02
Mostafa Ayaz
8,6723630
Look at the figure below
the curvature of $ax^4$ is always positive and that of $ln x$ is always negative (by twice differentiating them) therefore they intersect in exactly one point if they are also tangent in that point i.e.$$ln x=ax^4\dfrac1x=4ax^3$$which leads to $$ax^4=1over 4$$ and by substituting we have $$ln x=dfrac14$$which yields to $$x=e^1over 4$$and $$a=dfrac14e$$also the case $ale 0$ is trivial since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
answered Jul 14 at 18:02
Mostafa Ayaz
8,6723630
edited Jul 14 at 18:16
answered Jul 14 at 18:02
Mostafa Ayaz
8,6723630
answered Jul 14 at 18:02
Mostafa Ayaz
8,6723630
answered Jul 14 at 18:02
Mostafa Ayaz
8,6723630
8,6723630
What if $a le 0$?
– fleablood
Jul 14 at 18:11
This is the trivial case since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
– Mostafa Ayaz
Jul 14 at 18:15
add a comment |Â
What if $a le 0$?
– fleablood
Jul 14 at 18:11
This is the trivial case since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
– Mostafa Ayaz
Jul 14 at 18:15
What if $a le 0$?
– fleablood
Jul 14 at 18:11
What if $a le 0$?
– fleablood
Jul 14 at 18:11
This is the trivial case since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
– Mostafa Ayaz
Jul 14 at 18:15
This is the trivial case since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $ln x$ and $ax^4$ are strictly increasing and decreasing respectively.
– Mostafa Ayaz
Jul 14 at 18:15
add a comment |Â
up vote
1
down vote
Solution
Obviously, $x>0$. $ln x=ax^4$ could be rewritten as $$a=fracln xx^4.$$ We want to find the range of $a$ such that $y=a$ and $y=f(x)=dfracln xx^4$ intersect at only one point.
Notice that $$f'(x)=frac1-4ln xx^5.$$ Let $y'=0$. We have $x=e^frac14.$ Morover, $f'(x)>0$ when $x<e^frac14$ and $f'(x)<0$ when $x>e^frac14.$ Thus, we may claim that $f(x)$ increases over $(0,e^frac14]$ and decrease over $[e^frac14,+infty].$ Futher, $f(x) in (-infty,dfrac14e]$ and $f(x) in (dfrac14e,0)$. Here, you should notice $$limlimits_x to +inftyfracln xx^4=limlimits_x to +inftyfrac1/ x4x^3=limlimits_x to +inftyfrac14x^4=0.$$
Now, we may obtain that the range of $a$ we want is $(-infty,0]bigcapdfrac14e$. Otherwise, there exist either two or none intersetion.
answered Jul 14 at 18:01
mengdie1982
2,972216
Thank you very much. I actually understand the reason behind such exercises but I could not figure how to start it. I got confused because a was not separated from x and could not understand why. Thank you very much, again, and everyone else who has commented.
– Radu Gabriel
Jul 14 at 18:36
Not at all. please correct me if I'm wrong.
– mengdie1982
Jul 14 at 18:45
add a comment |Â
up vote
1
down vote
Solution
Obviously, $x>0$. $ln x=ax^4$ could be rewritten as $$a=fracln xx^4.$$ We want to find the range of $a$ such that $y=a$ and $y=f(x)=dfracln xx^4$ intersect at only one point.
Notice that $$f'(x)=frac1-4ln xx^5.$$ Let $y'=0$. We have $x=e^frac14.$ Morover, $f'(x)>0$ when $x<e^frac14$ and $f'(x)<0$ when $x>e^frac14.$ Thus, we may claim that $f(x)$ increases over $(0,e^frac14]$ and decrease over $[e^frac14,+infty].$ Futher, $f(x) in (-infty,dfrac14e]$ and $f(x) in (dfrac14e,0)$. Here, you should notice $$limlimits_x to +inftyfracln xx^4=limlimits_x to +inftyfrac1/ x4x^3=limlimits_x to +inftyfrac14x^4=0.$$
Now, we may obtain that the range of $a$ we want is $(-infty,0]bigcapdfrac14e$. Otherwise, there exist either two or none intersetion.
answered Jul 14 at 18:01
mengdie1982
2,972216
Thank you very much. I actually understand the reason behind such exercises but I could not figure how to start it. I got confused because a was not separated from x and could not understand why. Thank you very much, again, and everyone else who has commented.
– Radu Gabriel
Jul 14 at 18:36
Not at all. please correct me if I'm wrong.
– mengdie1982
Jul 14 at 18:45
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Solution
Obviously, $x>0$. $ln x=ax^4$ could be rewritten as $$a=fracln xx^4.$$ We want to find the range of $a$ such that $y=a$ and $y=f(x)=dfracln xx^4$ intersect at only one point.
Notice that $$f'(x)=frac1-4ln xx^5.$$ Let $y'=0$. We have $x=e^frac14.$ Morover, $f'(x)>0$ when $x<e^frac14$ and $f'(x)<0$ when $x>e^frac14.$ Thus, we may claim that $f(x)$ increases over $(0,e^frac14]$ and decrease over $[e^frac14,+infty].$ Futher, $f(x) in (-infty,dfrac14e]$ and $f(x) in (dfrac14e,0)$. Here, you should notice $$limlimits_x to +inftyfracln xx^4=limlimits_x to +inftyfrac1/ x4x^3=limlimits_x to +inftyfrac14x^4=0.$$
Now, we may obtain that the range of $a$ we want is $(-infty,0]bigcapdfrac14e$. Otherwise, there exist either two or none intersetion.
answered Jul 14 at 18:01
mengdie1982
2,972216
Solution
Obviously, $x>0$. $ln x=ax^4$ could be rewritten as $$a=fracln xx^4.$$ We want to find the range of $a$ such that $y=a$ and $y=f(x)=dfracln xx^4$ intersect at only one point.
Notice that $$f'(x)=frac1-4ln xx^5.$$ Let $y'=0$. We have $x=e^frac14.$ Morover, $f'(x)>0$ when $x<e^frac14$ and $f'(x)<0$ when $x>e^frac14.$ Thus, we may claim that $f(x)$ increases over $(0,e^frac14]$ and decrease over $[e^frac14,+infty].$ Futher, $f(x) in (-infty,dfrac14e]$ and $f(x) in (dfrac14e,0)$. Here, you should notice $$limlimits_x to +inftyfracln xx^4=limlimits_x to +inftyfrac1/ x4x^3=limlimits_x to +inftyfrac14x^4=0.$$
Now, we may obtain that the range of $a$ we want is $(-infty,0]bigcapdfrac14e$. Otherwise, there exist either two or none intersetion.
answered Jul 14 at 18:01
mengdie1982
2,972216
answered Jul 14 at 18:01
mengdie1982
2,972216
answered Jul 14 at 18:01
mengdie1982
2,972216
answered Jul 14 at 18:01
mengdie1982
2,972216
2,972216
Thank you very much. I actually understand the reason behind such exercises but I could not figure how to start it. I got confused because a was not separated from x and could not understand why. Thank you very much, again, and everyone else who has commented.
– Radu Gabriel
Jul 14 at 18:36
Not at all. please correct me if I'm wrong.
– mengdie1982
Jul 14 at 18:45
add a comment |Â
Thank you very much. I actually understand the reason behind such exercises but I could not figure how to start it. I got confused because a was not separated from x and could not understand why. Thank you very much, again, and everyone else who has commented.
– Radu Gabriel
Jul 14 at 18:36
Not at all. please correct me if I'm wrong.
– mengdie1982
Jul 14 at 18:45
Thank you very much. I actually understand the reason behind such exercises but I could not figure how to start it. I got confused because a was not separated from x and could not understand why. Thank you very much, again, and everyone else who has commented.
– Radu Gabriel
Jul 14 at 18:36
Thank you very much. I actually understand the reason behind such exercises but I could not figure how to start it. I got confused because a was not separated from x and could not understand why. Thank you very much, again, and everyone else who has commented.
– Radu Gabriel
Jul 14 at 18:36
Not at all. please correct me if I'm wrong.
– mengdie1982
Jul 14 at 18:45
Not at all. please correct me if I'm wrong.
– mengdie1982
Jul 14 at 18:45
add a comment |Â
up vote
0
down vote
It is easy to show that the function $ln x-ax^4$ has at most a single extremum, at $x=dfrac1sqrt[4]4a$.
For any $ale0$, there is no extremum and a single root, as the function goes from $-infty$ to $infty$.
If $a>0$, the value at the extremum is
$$frac14ln4a-frac14$$ and is zero for $4a=e$, forming a double root.
answered Jul 14 at 17:29
Yves Daoust
111k665205
add a comment |Â
up vote
0
down vote
It is easy to show that the function $ln x-ax^4$ has at most a single extremum, at $x=dfrac1sqrt[4]4a$.
For any $ale0$, there is no extremum and a single root, as the function goes from $-infty$ to $infty$.
If $a>0$, the value at the extremum is
$$frac14ln4a-frac14$$ and is zero for $4a=e$, forming a double root.
answered Jul 14 at 17:29
Yves Daoust
111k665205
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is easy to show that the function $ln x-ax^4$ has at most a single extremum, at $x=dfrac1sqrt[4]4a$.
For any $ale0$, there is no extremum and a single root, as the function goes from $-infty$ to $infty$.
If $a>0$, the value at the extremum is
$$frac14ln4a-frac14$$ and is zero for $4a=e$, forming a double root.
answered Jul 14 at 17:29
Yves Daoust
111k665205
It is easy to show that the function $ln x-ax^4$ has at most a single extremum, at $x=dfrac1sqrt[4]4a$.
For any $ale0$, there is no extremum and a single root, as the function goes from $-infty$ to $infty$.
If $a>0$, the value at the extremum is
$$frac14ln4a-frac14$$ and is zero for $4a=e$, forming a double root.
answered Jul 14 at 17:29
Yves Daoust
111k665205
answered Jul 14 at 17:29
Yves Daoust
111k665205
answered Jul 14 at 17:29
Yves Daoust
111k665205
answered Jul 14 at 17:29
Yves Daoust
111k665205
111k665205
add a comment |Â
add a comment |Â
up vote
0
down vote
Assume $a >0$.
Considering the U/valley shape of the graph of $ax^4$ and the slouching slug shape of $ln x$ it's clear that the graphs either never intersect and $ln x < ax^4$ for all $x> 0$. Or they touch once and there is one solution to $ln x = ax^4$ or the cross and recross and there are two solutions.
Now to solve $ln x = ax^4$ is to solve $h(x) = ax^4 - ln x = 0$ and as $a^4 ge ln x$ (the graphs touch once and never cross) then $h(x) = 0$ is local minimum.
So $h(x) = ax^4 - ln x = 0$ and $h'(x) = 4ax^3 - frac 1x =0$. Then later becomes $4ax^4 - 1 = 0$ and notice both $ax^4 - ln x = 0$ and $4ax^4 - 1=0$ have terms pertaining to $ax^4$. And we can solve $ax^4 = frac 14$.
So $frac 14 - ln x = 0$ and $ln x = frac 14$ and $x = e^frac 14$.
Plugging that in we get $a*e - frac 14 = 0$.
Which gives us $a = frac 14e$
If $a =0$ then $ln x = ax^4 = 0$ and $x = 0$ is the onlys solution.
If $a < 0$. Well, look at the graphs. The U valley is upside down and every value intersect the slouching slug exactly once. So $a in (-infty, 0]cup frac 14e$.
answered Jul 14 at 17:55
fleablood
60.5k22575
add a comment |Â
up vote
0
down vote
Assume $a >0$.
Considering the U/valley shape of the graph of $ax^4$ and the slouching slug shape of $ln x$ it's clear that the graphs either never intersect and $ln x < ax^4$ for all $x> 0$. Or they touch once and there is one solution to $ln x = ax^4$ or the cross and recross and there are two solutions.
Now to solve $ln x = ax^4$ is to solve $h(x) = ax^4 - ln x = 0$ and as $a^4 ge ln x$ (the graphs touch once and never cross) then $h(x) = 0$ is local minimum.
So $h(x) = ax^4 - ln x = 0$ and $h'(x) = 4ax^3 - frac 1x =0$. Then later becomes $4ax^4 - 1 = 0$ and notice both $ax^4 - ln x = 0$ and $4ax^4 - 1=0$ have terms pertaining to $ax^4$. And we can solve $ax^4 = frac 14$.
So $frac 14 - ln x = 0$ and $ln x = frac 14$ and $x = e^frac 14$.
Plugging that in we get $a*e - frac 14 = 0$.
Which gives us $a = frac 14e$
If $a =0$ then $ln x = ax^4 = 0$ and $x = 0$ is the onlys solution.
If $a < 0$. Well, look at the graphs. The U valley is upside down and every value intersect the slouching slug exactly once. So $a in (-infty, 0]cup frac 14e$.
answered Jul 14 at 17:55
fleablood
60.5k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assume $a >0$.
Considering the U/valley shape of the graph of $ax^4$ and the slouching slug shape of $ln x$ it's clear that the graphs either never intersect and $ln x < ax^4$ for all $x> 0$. Or they touch once and there is one solution to $ln x = ax^4$ or the cross and recross and there are two solutions.
Now to solve $ln x = ax^4$ is to solve $h(x) = ax^4 - ln x = 0$ and as $a^4 ge ln x$ (the graphs touch once and never cross) then $h(x) = 0$ is local minimum.
So $h(x) = ax^4 - ln x = 0$ and $h'(x) = 4ax^3 - frac 1x =0$. Then later becomes $4ax^4 - 1 = 0$ and notice both $ax^4 - ln x = 0$ and $4ax^4 - 1=0$ have terms pertaining to $ax^4$. And we can solve $ax^4 = frac 14$.
So $frac 14 - ln x = 0$ and $ln x = frac 14$ and $x = e^frac 14$.
Plugging that in we get $a*e - frac 14 = 0$.
Which gives us $a = frac 14e$
If $a =0$ then $ln x = ax^4 = 0$ and $x = 0$ is the onlys solution.
If $a < 0$. Well, look at the graphs. The U valley is upside down and every value intersect the slouching slug exactly once. So $a in (-infty, 0]cup frac 14e$.
answered Jul 14 at 17:55
fleablood
60.5k22575
Assume $a >0$.
Considering the U/valley shape of the graph of $ax^4$ and the slouching slug shape of $ln x$ it's clear that the graphs either never intersect and $ln x < ax^4$ for all $x> 0$. Or they touch once and there is one solution to $ln x = ax^4$ or the cross and recross and there are two solutions.
Now to solve $ln x = ax^4$ is to solve $h(x) = ax^4 - ln x = 0$ and as $a^4 ge ln x$ (the graphs touch once and never cross) then $h(x) = 0$ is local minimum.
So $h(x) = ax^4 - ln x = 0$ and $h'(x) = 4ax^3 - frac 1x =0$. Then later becomes $4ax^4 - 1 = 0$ and notice both $ax^4 - ln x = 0$ and $4ax^4 - 1=0$ have terms pertaining to $ax^4$. And we can solve $ax^4 = frac 14$.
So $frac 14 - ln x = 0$ and $ln x = frac 14$ and $x = e^frac 14$.
Plugging that in we get $a*e - frac 14 = 0$.
Which gives us $a = frac 14e$
If $a =0$ then $ln x = ax^4 = 0$ and $x = 0$ is the onlys solution.
If $a < 0$. Well, look at the graphs. The U valley is upside down and every value intersect the slouching slug exactly once. So $a in (-infty, 0]cup frac 14e$.
answered Jul 14 at 17:55
fleablood
60.5k22575
edited Jul 14 at 18:10
answered Jul 14 at 17:55
fleablood
60.5k22575
answered Jul 14 at 17:55
fleablood
60.5k22575
answered Jul 14 at 17:55
fleablood
60.5k22575
60.5k22575
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint: Consider the function
$f(x)=fracln(x)x^4$ and use calculus.
Use that
$$f'(x)=frac1-4ln(x)x^4$$
answered Jul 14 at 17:13
Dr. Sonnhard Graubner
67k32660
1
don't you mean $f'(x)=frac1-4ln(x)x^5$
– Bruce
Jul 14 at 17:17
Thank you for your hint!
– Dr. Sonnhard Graubner
Jul 14 at 18:21
add a comment |Â
up vote
0
down vote
Hint: Consider the function
$f(x)=fracln(x)x^4$ and use calculus.
Use that
$$f'(x)=frac1-4ln(x)x^4$$
answered Jul 14 at 17:13
Dr. Sonnhard Graubner
67k32660
1
don't you mean $f'(x)=frac1-4ln(x)x^5$
– Bruce
Jul 14 at 17:17
Thank you for your hint!
– Dr. Sonnhard Graubner
Jul 14 at 18:21
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: Consider the function
$f(x)=fracln(x)x^4$ and use calculus.
Use that
$$f'(x)=frac1-4ln(x)x^4$$
answered Jul 14 at 17:13
Dr. Sonnhard Graubner
67k32660
Hint: Consider the function
$f(x)=fracln(x)x^4$ and use calculus.
Use that
$$f'(x)=frac1-4ln(x)x^4$$
answered Jul 14 at 17:13
Dr. Sonnhard Graubner
67k32660
edited Jul 14 at 18:20
answered Jul 14 at 17:13
Dr. Sonnhard Graubner
67k32660
answered Jul 14 at 17:13
Dr. Sonnhard Graubner
67k32660
answered Jul 14 at 17:13
Dr. Sonnhard Graubner
67k32660
67k32660
1
don't you mean $f'(x)=frac1-4ln(x)x^5$
– Bruce
Jul 14 at 17:17
Thank you for your hint!
– Dr. Sonnhard Graubner
Jul 14 at 18:21
add a comment |Â
1
don't you mean $f'(x)=frac1-4ln(x)x^5$
– Bruce
Jul 14 at 17:17
Thank you for your hint!
– Dr. Sonnhard Graubner
Jul 14 at 18:21
1
1
don't you mean $f'(x)=frac1-4ln(x)x^5$
– Bruce
Jul 14 at 17:17
don't you mean $f'(x)=frac1-4ln(x)x^5$
– Bruce
Jul 14 at 17:17
Thank you for your hint!
– Dr. Sonnhard Graubner
Jul 14 at 18:21
Thank you for your hint!
– Dr. Sonnhard Graubner
Jul 14 at 18:21
add a comment |Â
up vote
0
down vote
beginalign
ln x&=ax^4
tag1label1
,\
-4ln x&=-4ax^4
,\
ln(x^-4)&=-4ax^4
,\
x^-4ln(x^-4)&=-4a
.
endalign
Let $ln(x^-4)=u$, then
$x^-4=exp(u)$
and we arrive at an equation
beginalign
uexp(u)&=-4a
,
endalign
which has a well-known
closed-form solution in terms of
Lambert W function.
beginalign
operatornameW(uexp(u))&=operatornameW(-4a)
,\
u&=operatornameW(-4a)
,\
ln(x^-4)&=operatornameW(-4a)
,\
x^-4&=exp(operatornameW(-4a))
,\
x&=exp(-tfrac14cdotoperatornameW(-4a))
tag2label2
.
endalign
The number of real solutions to eqref1
is defined by the number of
real solutions to eqref2,
which in turn, is defined by the argument of $operatornameW$,
and it is well-known
that we have a single real solution
when the argument of $operatornameW$
is either non-negative, or is equal to
$-tfrac1mathrme$.
That is,
beginalign
1)& -4age0quadtoquad ale0
;\
2)& -4a=-tfrac1mathrme
quadtoquad
a=tfrac1mathrm4e
approx 0.09196986
.
endalign
In the last case we have
beginalign
x&=exp(-tfrac14cdotoperatornameW(-tfrac1mathrme))
=exp(-tfrac14cdot(-1))
=sqrt[4]mathrme
.
endalign
answered Jul 14 at 19:33
g.kov
5,5521717
add a comment |Â
up vote
0
down vote
beginalign
ln x&=ax^4
tag1label1
,\
-4ln x&=-4ax^4
,\
ln(x^-4)&=-4ax^4
,\
x^-4ln(x^-4)&=-4a
.
endalign
Let $ln(x^-4)=u$, then
$x^-4=exp(u)$
and we arrive at an equation
beginalign
uexp(u)&=-4a
,
endalign
which has a well-known
closed-form solution in terms of
Lambert W function.
beginalign
operatornameW(uexp(u))&=operatornameW(-4a)
,\
u&=operatornameW(-4a)
,\
ln(x^-4)&=operatornameW(-4a)
,\
x^-4&=exp(operatornameW(-4a))
,\
x&=exp(-tfrac14cdotoperatornameW(-4a))
tag2label2
.
endalign
The number of real solutions to eqref1
is defined by the number of
real solutions to eqref2,
which in turn, is defined by the argument of $operatornameW$,
and it is well-known
that we have a single real solution
when the argument of $operatornameW$
is either non-negative, or is equal to
$-tfrac1mathrme$.
That is,
beginalign
1)& -4age0quadtoquad ale0
;\
2)& -4a=-tfrac1mathrme
quadtoquad
a=tfrac1mathrm4e
approx 0.09196986
.
endalign
In the last case we have
beginalign
x&=exp(-tfrac14cdotoperatornameW(-tfrac1mathrme))
=exp(-tfrac14cdot(-1))
=sqrt[4]mathrme
.
endalign
answered Jul 14 at 19:33
g.kov
5,5521717
add a comment |Â
up vote
0
down vote
up vote
0
down vote
beginalign
ln x&=ax^4
tag1label1
,\
-4ln x&=-4ax^4
,\
ln(x^-4)&=-4ax^4
,\
x^-4ln(x^-4)&=-4a
.
endalign
Let $ln(x^-4)=u$, then
$x^-4=exp(u)$
and we arrive at an equation
beginalign
uexp(u)&=-4a
,
endalign
which has a well-known
closed-form solution in terms of
Lambert W function.
beginalign
operatornameW(uexp(u))&=operatornameW(-4a)
,\
u&=operatornameW(-4a)
,\
ln(x^-4)&=operatornameW(-4a)
,\
x^-4&=exp(operatornameW(-4a))
,\
x&=exp(-tfrac14cdotoperatornameW(-4a))
tag2label2
.
endalign
The number of real solutions to eqref1
is defined by the number of
real solutions to eqref2,
which in turn, is defined by the argument of $operatornameW$,
and it is well-known
that we have a single real solution
when the argument of $operatornameW$
is either non-negative, or is equal to
$-tfrac1mathrme$.
That is,
beginalign
1)& -4age0quadtoquad ale0
;\
2)& -4a=-tfrac1mathrme
quadtoquad
a=tfrac1mathrm4e
approx 0.09196986
.
endalign
In the last case we have
beginalign
x&=exp(-tfrac14cdotoperatornameW(-tfrac1mathrme))
=exp(-tfrac14cdot(-1))
=sqrt[4]mathrme
.
endalign
answered Jul 14 at 19:33
g.kov
5,5521717
beginalign
ln x&=ax^4
tag1label1
,\
-4ln x&=-4ax^4
,\
ln(x^-4)&=-4ax^4
,\
x^-4ln(x^-4)&=-4a
.
endalign
Let $ln(x^-4)=u$, then
$x^-4=exp(u)$
and we arrive at an equation
beginalign
uexp(u)&=-4a
,
endalign
which has a well-known
closed-form solution in terms of
Lambert W function.
beginalign
operatornameW(uexp(u))&=operatornameW(-4a)
,\
u&=operatornameW(-4a)
,\
ln(x^-4)&=operatornameW(-4a)
,\
x^-4&=exp(operatornameW(-4a))
,\
x&=exp(-tfrac14cdotoperatornameW(-4a))
tag2label2
.
endalign
The number of real solutions to eqref1
is defined by the number of
real solutions to eqref2,
which in turn, is defined by the argument of $operatornameW$,
and it is well-known
that we have a single real solution
when the argument of $operatornameW$
is either non-negative, or is equal to
$-tfrac1mathrme$.
That is,
beginalign
1)& -4age0quadtoquad ale0
;\
2)& -4a=-tfrac1mathrme
quadtoquad
a=tfrac1mathrm4e
approx 0.09196986
.
endalign
In the last case we have
beginalign
x&=exp(-tfrac14cdotoperatornameW(-tfrac1mathrme))
=exp(-tfrac14cdot(-1))
=sqrt[4]mathrme
.
endalign
answered Jul 14 at 19:33
g.kov
5,5521717
answered Jul 14 at 19:33
g.kov
5,5521717
answered Jul 14 at 19:33
g.kov
5,5521717
answered Jul 14 at 19:33
g.kov
5,5521717
5,5521717
add a comment |Â
add a comment |Â
Are we assuming $a > 0$. Other wise $a = 0$ will yield $ln x = 0$ has only one solution.
– fleablood
Jul 14 at 18:10