Can one solve for $x$ in the equation $8^x=16x$ [closed]

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The question is whether or not it is possible to solve for $x$ in the equation $8^x=16x$.

[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/ypio1.jpg any assistance will be appreciated

logarithms exponential-function

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edited Jul 30 at 7:11

MathIsHard

1,122415

asked Jul 30 at 6:57

user451954

6

closed as off-topic by amWhy, Rhys Steele, Mostafa Ayaz, Isaac Browne, Strants Jul 30 at 20:59

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." – amWhy, Rhys Steele, Mostafa Ayaz, Isaac Browne, Strants

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please type out the question in MathJax, and then show your own thoughts and work. What is the difficulty you are facing?
  – Matti P.
  Jul 30 at 7:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Quick answer: I don't think there is an algebraic (exact) solution, but it's relatively easy to find an approximation.
  – Matti P.
  Jul 30 at 7:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Lookup the Lambert W function.
  – dxiv
  Jul 30 at 7:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E.g. wolframalpha.com/input/?i=8%5Ex%3D16x ; then press Approximate forms to get the numerical solutions
  – user90369
  Jul 30 at 7:22
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

The question is whether or not it is possible to solve for $x$ in the equation $8^x=16x$.

[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/ypio1.jpg any assistance will be appreciated

logarithms exponential-function

share|cite|improve this question

edited Jul 30 at 7:11

MathIsHard

1,122415

asked Jul 30 at 6:57

user451954

6

closed as off-topic by amWhy, Rhys Steele, Mostafa Ayaz, Isaac Browne, Strants Jul 30 at 20:59

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." – amWhy, Rhys Steele, Mostafa Ayaz, Isaac Browne, Strants

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please type out the question in MathJax, and then show your own thoughts and work. What is the difficulty you are facing?
  – Matti P.
  Jul 30 at 7:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Quick answer: I don't think there is an algebraic (exact) solution, but it's relatively easy to find an approximation.
  – Matti P.
  Jul 30 at 7:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Lookup the Lambert W function.
  – dxiv
  Jul 30 at 7:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E.g. wolframalpha.com/input/?i=8%5Ex%3D16x ; then press Approximate forms to get the numerical solutions
  – user90369
  Jul 30 at 7:22
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

The question is whether or not it is possible to solve for $x$ in the equation $8^x=16x$.

[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/ypio1.jpg any assistance will be appreciated

logarithms exponential-function

share|cite|improve this question

edited Jul 30 at 7:11

MathIsHard

1,122415

asked Jul 30 at 6:57

user451954

6

The question is whether or not it is possible to solve for $x$ in the equation $8^x=16x$.

[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/ypio1.jpg any assistance will be appreciated

logarithms exponential-function

share|cite|improve this question

edited Jul 30 at 7:11

MathIsHard

1,122415

asked Jul 30 at 6:57

user451954

6

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 7:11

MathIsHard

1,122415

edited Jul 30 at 7:11

MathIsHard

1,122415

edited Jul 30 at 7:11

MathIsHard

1,122415

1,122415

asked Jul 30 at 6:57

user451954

6

asked Jul 30 at 6:57

user451954

6

asked Jul 30 at 6:57

user451954

6

6

closed as off-topic by amWhy, Rhys Steele, Mostafa Ayaz, Isaac Browne, Strants Jul 30 at 20:59

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." – amWhy, Rhys Steele, Mostafa Ayaz, Isaac Browne, Strants

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Rhys Steele, Mostafa Ayaz, Isaac Browne, Strants Jul 30 at 20:59

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." – amWhy, Rhys Steele, Mostafa Ayaz, Isaac Browne, Strants

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please type out the question in MathJax, and then show your own thoughts and work. What is the difficulty you are facing?
  – Matti P.
  Jul 30 at 7:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Quick answer: I don't think there is an algebraic (exact) solution, but it's relatively easy to find an approximation.
  – Matti P.
  Jul 30 at 7:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Lookup the Lambert W function.
  – dxiv
  Jul 30 at 7:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E.g. wolframalpha.com/input/?i=8%5Ex%3D16x ; then press Approximate forms to get the numerical solutions
  – user90369
  Jul 30 at 7:22
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please type out the question in MathJax, and then show your own thoughts and work. What is the difficulty you are facing?
  – Matti P.
  Jul 30 at 7:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Quick answer: I don't think there is an algebraic (exact) solution, but it's relatively easy to find an approximation.
  – Matti P.
  Jul 30 at 7:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Lookup the Lambert W function.
  – dxiv
  Jul 30 at 7:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E.g. wolframalpha.com/input/?i=8%5Ex%3D16x ; then press Approximate forms to get the numerical solutions
  – user90369
  Jul 30 at 7:22
  
  

  
  
  
  

Please type out the question in MathJax, and then show your own thoughts and work. What is the difficulty you are facing?
– Matti P.
Jul 30 at 7:03

Please type out the question in MathJax, and then show your own thoughts and work. What is the difficulty you are facing?
– Matti P.
Jul 30 at 7:03

2

2

Quick answer: I don't think there is an algebraic (exact) solution, but it's relatively easy to find an approximation.
– Matti P.
Jul 30 at 7:04

Quick answer: I don't think there is an algebraic (exact) solution, but it's relatively easy to find an approximation.
– Matti P.
Jul 30 at 7:04

1

1

Lookup the Lambert W function.
– dxiv
Jul 30 at 7:14

Lookup the Lambert W function.
– dxiv
Jul 30 at 7:14

E.g. wolframalpha.com/input/?i=8%5Ex%3D16x ; then press Approximate forms to get the numerical solutions
– user90369
Jul 30 at 7:22

E.g. wolframalpha.com/input/?i=8%5Ex%3D16x ; then press Approximate forms to get the numerical solutions
– user90369
Jul 30 at 7:22

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote

$$8^x=16x$$
$$e^xln(8)=16x$$
$$e^-xln(8)=frac116x$$
$$xe^-xln(8)=frac116$$
$$-xln(8)e^-xln(8)= -frac116ln(8)$$
Let $quad z=-xln(8)quad$ and $quad C=-frac116ln(8)$
$$ze^z=C$$
This is the usual Lambert's equation which solutions are :
$$z=W(C)$$
$W$ is the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html

$$x=-fraczln(8)= -frac1ln(8)Wleft(Cright)$$
$$x=-frac1ln(8)Wleft(-frac116ln(8)right)$$

The Lambert W function is multi-valuated. The real solutions are commonly noted $W_0(X)$ and $W_-1(X)$. Finally the two real solutions are :
$$x=-frac1ln(8)W_0left(-frac116ln(8)right)simeq 0.0727$$
$$x=-frac1ln(8)W_-1left(-frac116ln(8)right)simeq 1.54142$$
http://m.wolframalpha.com/input/?i=-lambertw%280%2C-ln%288%29%2F16%29%2Fln%288%29

http://m.wolframalpha.com/input/?i=-lambertw%28-1%2C-ln%288%29%2F16%29%2Fln%288%29

Of course, instead of using this special function, in practice the numerical approximate solutions are directly computed thanks to numerical method of calculus (Newton-Raphson for example).

share|cite|improve this answer

edited Jul 30 at 8:39

answered Jul 30 at 8:28

JJacquelin

39.7k21649

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice! How would one denote the real solution if there were more than two?
  – Zacky
  Jul 30 at 8:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Never the W function has more than two real solutions. The whole set of solutions (complex and real) is denoted $W_n(X)$.
  – JJacquelin
  Jul 30 at 9:13
  

  
  
  
  

add a comment | 

up vote
1
down vote

Well, the most general problem is (when $textcne1$):

$$texta+textbcdottextc^x=textdcdot xspaceLongleftrightarrowspace x=fractextatextd-fractextW_textnleft(-fractextbtextdcdottextc^fractextatextdcdotlnleft(textcright)right)lnleft(textcright)tag1$$

Where $textW_textkleft(textzright)$ is the analytic continuation of the Product Log Function and $textninmathbbZ$.

---

So, in your problem:

$$0+1cdot8^x=16cdot xspaceLongleftrightarrowspace x=frac016-fractextW_textnleft(-frac116cdot8^frac016cdotlnleft(8right)right)lnleft(8right)=-fractextW_textnleft(-fraclnleft(8right)16right)lnleft(8right)tag2$$

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answered Jul 30 at 9:20

Jan

21.6k31239

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up vote
0
down vote

The solution given using Lambert’s function is very elegant. However, if you are interested in obtaining the solution using numerical methods, you can follow this:

Let $f(x) = 8^x-16x$. This a smooth function.

It's obvious from the graph of $f(x)$ that it cuts x-axis, and hence you are guaranteed a solution. In fact, you can see it has two solutions.

Otherwise, use IVT to verify this:
$f(1) = -6$ and $f(0) = 1$ and $f(1) = -6$ and $f(2) > 0$

By IVT, you are guaranteed a point $x_sol in (0,1)$ and one more in $(1,2)$ such that $f(x_sol)=0$. Since the function is monotonic ouside $(0,1)$ you will not get anymore solution points. )

You can use Newton's method to find the solution $x_sol$ in (0,1) as follows:

Choose an initial point $x_0$ for Newton's method in $(0,1)$.

$df(x) = 8^x ln 8-16$.

$|df(a)-df(b)| leq ln 8 |8^a-8^b|$ $forall$ $a,b in (0,1)$,

$implies |df(a)-df(b)| leq ln 8 |a-b|$

Thus, $df(x)$ is Lipschitz in $(0,1)$ with a Lipschitz ratio $M = ln 8$ in $(0,1)$,

Let $x_0 in (0,1)$

You can show that, $|f(x_0)M df^-1(x_0)^2| leq 0.5$

and hence you are guaranteed to converge to a solution $x_sol in (0,1)$ by Katorovich's theorem.

If you want the other solution point in $(1,2)$, choose $x_0 in (1,2)$, and Kantorovich’s theorem guarantees convergence of Newton’s method to $x_sol in (1,2)$

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answered Jul 30 at 11:16

Suhan Shetty

835

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

$$8^x=16x$$
$$e^xln(8)=16x$$
$$e^-xln(8)=frac116x$$
$$xe^-xln(8)=frac116$$
$$-xln(8)e^-xln(8)= -frac116ln(8)$$
Let $quad z=-xln(8)quad$ and $quad C=-frac116ln(8)$
$$ze^z=C$$
This is the usual Lambert's equation which solutions are :
$$z=W(C)$$
$W$ is the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html

$$x=-fraczln(8)= -frac1ln(8)Wleft(Cright)$$
$$x=-frac1ln(8)Wleft(-frac116ln(8)right)$$

The Lambert W function is multi-valuated. The real solutions are commonly noted $W_0(X)$ and $W_-1(X)$. Finally the two real solutions are :
$$x=-frac1ln(8)W_0left(-frac116ln(8)right)simeq 0.0727$$
$$x=-frac1ln(8)W_-1left(-frac116ln(8)right)simeq 1.54142$$
http://m.wolframalpha.com/input/?i=-lambertw%280%2C-ln%288%29%2F16%29%2Fln%288%29

http://m.wolframalpha.com/input/?i=-lambertw%28-1%2C-ln%288%29%2F16%29%2Fln%288%29

Of course, instead of using this special function, in practice the numerical approximate solutions are directly computed thanks to numerical method of calculus (Newton-Raphson for example).

share|cite|improve this answer

edited Jul 30 at 8:39

answered Jul 30 at 8:28

JJacquelin

39.7k21649

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice! How would one denote the real solution if there were more than two?
  – Zacky
  Jul 30 at 8:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Never the W function has more than two real solutions. The whole set of solutions (complex and real) is denoted $W_n(X)$.
  – JJacquelin
  Jul 30 at 9:13
  

  
  
  
  

add a comment | 

up vote
2
down vote

$$8^x=16x$$
$$e^xln(8)=16x$$
$$e^-xln(8)=frac116x$$
$$xe^-xln(8)=frac116$$
$$-xln(8)e^-xln(8)= -frac116ln(8)$$
Let $quad z=-xln(8)quad$ and $quad C=-frac116ln(8)$
$$ze^z=C$$
This is the usual Lambert's equation which solutions are :
$$z=W(C)$$
$W$ is the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html

$$x=-fraczln(8)= -frac1ln(8)Wleft(Cright)$$
$$x=-frac1ln(8)Wleft(-frac116ln(8)right)$$

The Lambert W function is multi-valuated. The real solutions are commonly noted $W_0(X)$ and $W_-1(X)$. Finally the two real solutions are :
$$x=-frac1ln(8)W_0left(-frac116ln(8)right)simeq 0.0727$$
$$x=-frac1ln(8)W_-1left(-frac116ln(8)right)simeq 1.54142$$
http://m.wolframalpha.com/input/?i=-lambertw%280%2C-ln%288%29%2F16%29%2Fln%288%29

http://m.wolframalpha.com/input/?i=-lambertw%28-1%2C-ln%288%29%2F16%29%2Fln%288%29

Of course, instead of using this special function, in practice the numerical approximate solutions are directly computed thanks to numerical method of calculus (Newton-Raphson for example).

share|cite|improve this answer

edited Jul 30 at 8:39

answered Jul 30 at 8:28

JJacquelin

39.7k21649

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice! How would one denote the real solution if there were more than two?
  – Zacky
  Jul 30 at 8:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Never the W function has more than two real solutions. The whole set of solutions (complex and real) is denoted $W_n(X)$.
  – JJacquelin
  Jul 30 at 9:13
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

$$8^x=16x$$
$$e^xln(8)=16x$$
$$e^-xln(8)=frac116x$$
$$xe^-xln(8)=frac116$$
$$-xln(8)e^-xln(8)= -frac116ln(8)$$
Let $quad z=-xln(8)quad$ and $quad C=-frac116ln(8)$
$$ze^z=C$$
This is the usual Lambert's equation which solutions are :
$$z=W(C)$$
$W$ is the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html

$$x=-fraczln(8)= -frac1ln(8)Wleft(Cright)$$
$$x=-frac1ln(8)Wleft(-frac116ln(8)right)$$

The Lambert W function is multi-valuated. The real solutions are commonly noted $W_0(X)$ and $W_-1(X)$. Finally the two real solutions are :
$$x=-frac1ln(8)W_0left(-frac116ln(8)right)simeq 0.0727$$
$$x=-frac1ln(8)W_-1left(-frac116ln(8)right)simeq 1.54142$$
http://m.wolframalpha.com/input/?i=-lambertw%280%2C-ln%288%29%2F16%29%2Fln%288%29

http://m.wolframalpha.com/input/?i=-lambertw%28-1%2C-ln%288%29%2F16%29%2Fln%288%29

Of course, instead of using this special function, in practice the numerical approximate solutions are directly computed thanks to numerical method of calculus (Newton-Raphson for example).

share|cite|improve this answer

edited Jul 30 at 8:39

answered Jul 30 at 8:28

JJacquelin

39.7k21649

$$8^x=16x$$
$$e^xln(8)=16x$$
$$e^-xln(8)=frac116x$$
$$xe^-xln(8)=frac116$$
$$-xln(8)e^-xln(8)= -frac116ln(8)$$
Let $quad z=-xln(8)quad$ and $quad C=-frac116ln(8)$
$$ze^z=C$$
This is the usual Lambert's equation which solutions are :
$$z=W(C)$$
$W$ is the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html

$$x=-fraczln(8)= -frac1ln(8)Wleft(Cright)$$
$$x=-frac1ln(8)Wleft(-frac116ln(8)right)$$

The Lambert W function is multi-valuated. The real solutions are commonly noted $W_0(X)$ and $W_-1(X)$. Finally the two real solutions are :
$$x=-frac1ln(8)W_0left(-frac116ln(8)right)simeq 0.0727$$
$$x=-frac1ln(8)W_-1left(-frac116ln(8)right)simeq 1.54142$$
http://m.wolframalpha.com/input/?i=-lambertw%280%2C-ln%288%29%2F16%29%2Fln%288%29

http://m.wolframalpha.com/input/?i=-lambertw%28-1%2C-ln%288%29%2F16%29%2Fln%288%29

Of course, instead of using this special function, in practice the numerical approximate solutions are directly computed thanks to numerical method of calculus (Newton-Raphson for example).

share|cite|improve this answer

edited Jul 30 at 8:39

answered Jul 30 at 8:28

JJacquelin

39.7k21649

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 8:39

edited Jul 30 at 8:39

edited Jul 30 at 8:39

answered Jul 30 at 8:28

JJacquelin

39.7k21649

answered Jul 30 at 8:28

JJacquelin

39.7k21649

answered Jul 30 at 8:28

JJacquelin

39.7k21649

39.7k21649

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice! How would one denote the real solution if there were more than two?
  – Zacky
  Jul 30 at 8:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Never the W function has more than two real solutions. The whole set of solutions (complex and real) is denoted $W_n(X)$.
  – JJacquelin
  Jul 30 at 9:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice! How would one denote the real solution if there were more than two?
  – Zacky
  Jul 30 at 8:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Never the W function has more than two real solutions. The whole set of solutions (complex and real) is denoted $W_n(X)$.
  – JJacquelin
  Jul 30 at 9:13
  

  
  
  
  

Nice! How would one denote the real solution if there were more than two?
– Zacky
Jul 30 at 8:44

Nice! How would one denote the real solution if there were more than two?
– Zacky
Jul 30 at 8:44

Never the W function has more than two real solutions. The whole set of solutions (complex and real) is denoted $W_n(X)$.
– JJacquelin
Jul 30 at 9:13

Never the W function has more than two real solutions. The whole set of solutions (complex and real) is denoted $W_n(X)$.
– JJacquelin
Jul 30 at 9:13

add a comment | 

up vote
1
down vote

Well, the most general problem is (when $textcne1$):

$$texta+textbcdottextc^x=textdcdot xspaceLongleftrightarrowspace x=fractextatextd-fractextW_textnleft(-fractextbtextdcdottextc^fractextatextdcdotlnleft(textcright)right)lnleft(textcright)tag1$$

Where $textW_textkleft(textzright)$ is the analytic continuation of the Product Log Function and $textninmathbbZ$.

---

So, in your problem:

$$0+1cdot8^x=16cdot xspaceLongleftrightarrowspace x=frac016-fractextW_textnleft(-frac116cdot8^frac016cdotlnleft(8right)right)lnleft(8right)=-fractextW_textnleft(-fraclnleft(8right)16right)lnleft(8right)tag2$$

share|cite|improve this answer

answered Jul 30 at 9:20

Jan

21.6k31239

add a comment | 

up vote
1
down vote

Well, the most general problem is (when $textcne1$):

$$texta+textbcdottextc^x=textdcdot xspaceLongleftrightarrowspace x=fractextatextd-fractextW_textnleft(-fractextbtextdcdottextc^fractextatextdcdotlnleft(textcright)right)lnleft(textcright)tag1$$

Where $textW_textkleft(textzright)$ is the analytic continuation of the Product Log Function and $textninmathbbZ$.

---

So, in your problem:

$$0+1cdot8^x=16cdot xspaceLongleftrightarrowspace x=frac016-fractextW_textnleft(-frac116cdot8^frac016cdotlnleft(8right)right)lnleft(8right)=-fractextW_textnleft(-fraclnleft(8right)16right)lnleft(8right)tag2$$

share|cite|improve this answer

answered Jul 30 at 9:20

Jan

21.6k31239

add a comment | 

up vote
1
down vote

up vote
1
down vote

Well, the most general problem is (when $textcne1$):

$$texta+textbcdottextc^x=textdcdot xspaceLongleftrightarrowspace x=fractextatextd-fractextW_textnleft(-fractextbtextdcdottextc^fractextatextdcdotlnleft(textcright)right)lnleft(textcright)tag1$$

Where $textW_textkleft(textzright)$ is the analytic continuation of the Product Log Function and $textninmathbbZ$.

---

So, in your problem:

$$0+1cdot8^x=16cdot xspaceLongleftrightarrowspace x=frac016-fractextW_textnleft(-frac116cdot8^frac016cdotlnleft(8right)right)lnleft(8right)=-fractextW_textnleft(-fraclnleft(8right)16right)lnleft(8right)tag2$$

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answered Jul 30 at 9:20

Jan

21.6k31239

Well, the most general problem is (when $textcne1$):

$$texta+textbcdottextc^x=textdcdot xspaceLongleftrightarrowspace x=fractextatextd-fractextW_textnleft(-fractextbtextdcdottextc^fractextatextdcdotlnleft(textcright)right)lnleft(textcright)tag1$$

Where $textW_textkleft(textzright)$ is the analytic continuation of the Product Log Function and $textninmathbbZ$.

---

So, in your problem:

$$0+1cdot8^x=16cdot xspaceLongleftrightarrowspace x=frac016-fractextW_textnleft(-frac116cdot8^frac016cdotlnleft(8right)right)lnleft(8right)=-fractextW_textnleft(-fraclnleft(8right)16right)lnleft(8right)tag2$$

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answered Jul 30 at 9:20

Jan

21.6k31239

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 9:20

Jan

21.6k31239

answered Jul 30 at 9:20

Jan

21.6k31239

answered Jul 30 at 9:20

Jan

21.6k31239

21.6k31239

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add a comment | 

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The solution given using Lambert’s function is very elegant. However, if you are interested in obtaining the solution using numerical methods, you can follow this:

Let $f(x) = 8^x-16x$. This a smooth function.

It's obvious from the graph of $f(x)$ that it cuts x-axis, and hence you are guaranteed a solution. In fact, you can see it has two solutions.

Otherwise, use IVT to verify this:
$f(1) = -6$ and $f(0) = 1$ and $f(1) = -6$ and $f(2) > 0$

By IVT, you are guaranteed a point $x_sol in (0,1)$ and one more in $(1,2)$ such that $f(x_sol)=0$. Since the function is monotonic ouside $(0,1)$ you will not get anymore solution points. )

You can use Newton's method to find the solution $x_sol$ in (0,1) as follows:

Choose an initial point $x_0$ for Newton's method in $(0,1)$.

$df(x) = 8^x ln 8-16$.

$|df(a)-df(b)| leq ln 8 |8^a-8^b|$ $forall$ $a,b in (0,1)$,

$implies |df(a)-df(b)| leq ln 8 |a-b|$

Thus, $df(x)$ is Lipschitz in $(0,1)$ with a Lipschitz ratio $M = ln 8$ in $(0,1)$,

Let $x_0 in (0,1)$

You can show that, $|f(x_0)M df^-1(x_0)^2| leq 0.5$

and hence you are guaranteed to converge to a solution $x_sol in (0,1)$ by Katorovich's theorem.

If you want the other solution point in $(1,2)$, choose $x_0 in (1,2)$, and Kantorovich’s theorem guarantees convergence of Newton’s method to $x_sol in (1,2)$

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answered Jul 30 at 11:16

Suhan Shetty

835

add a comment | 

up vote
0
down vote

The solution given using Lambert’s function is very elegant. However, if you are interested in obtaining the solution using numerical methods, you can follow this:

Let $f(x) = 8^x-16x$. This a smooth function.

It's obvious from the graph of $f(x)$ that it cuts x-axis, and hence you are guaranteed a solution. In fact, you can see it has two solutions.

Otherwise, use IVT to verify this:
$f(1) = -6$ and $f(0) = 1$ and $f(1) = -6$ and $f(2) > 0$

By IVT, you are guaranteed a point $x_sol in (0,1)$ and one more in $(1,2)$ such that $f(x_sol)=0$. Since the function is monotonic ouside $(0,1)$ you will not get anymore solution points. )

You can use Newton's method to find the solution $x_sol$ in (0,1) as follows:

Choose an initial point $x_0$ for Newton's method in $(0,1)$.

$df(x) = 8^x ln 8-16$.

$|df(a)-df(b)| leq ln 8 |8^a-8^b|$ $forall$ $a,b in (0,1)$,

$implies |df(a)-df(b)| leq ln 8 |a-b|$

Thus, $df(x)$ is Lipschitz in $(0,1)$ with a Lipschitz ratio $M = ln 8$ in $(0,1)$,

Let $x_0 in (0,1)$

You can show that, $|f(x_0)M df^-1(x_0)^2| leq 0.5$

and hence you are guaranteed to converge to a solution $x_sol in (0,1)$ by Katorovich's theorem.

If you want the other solution point in $(1,2)$, choose $x_0 in (1,2)$, and Kantorovich’s theorem guarantees convergence of Newton’s method to $x_sol in (1,2)$

share|cite|improve this answer

answered Jul 30 at 11:16

Suhan Shetty

835

add a comment | 

up vote
0
down vote

up vote
0
down vote

The solution given using Lambert’s function is very elegant. However, if you are interested in obtaining the solution using numerical methods, you can follow this:

Let $f(x) = 8^x-16x$. This a smooth function.

It's obvious from the graph of $f(x)$ that it cuts x-axis, and hence you are guaranteed a solution. In fact, you can see it has two solutions.

Otherwise, use IVT to verify this:
$f(1) = -6$ and $f(0) = 1$ and $f(1) = -6$ and $f(2) > 0$

By IVT, you are guaranteed a point $x_sol in (0,1)$ and one more in $(1,2)$ such that $f(x_sol)=0$. Since the function is monotonic ouside $(0,1)$ you will not get anymore solution points. )

You can use Newton's method to find the solution $x_sol$ in (0,1) as follows:

Choose an initial point $x_0$ for Newton's method in $(0,1)$.

$df(x) = 8^x ln 8-16$.

$|df(a)-df(b)| leq ln 8 |8^a-8^b|$ $forall$ $a,b in (0,1)$,

$implies |df(a)-df(b)| leq ln 8 |a-b|$

Thus, $df(x)$ is Lipschitz in $(0,1)$ with a Lipschitz ratio $M = ln 8$ in $(0,1)$,

Let $x_0 in (0,1)$

You can show that, $|f(x_0)M df^-1(x_0)^2| leq 0.5$

and hence you are guaranteed to converge to a solution $x_sol in (0,1)$ by Katorovich's theorem.

If you want the other solution point in $(1,2)$, choose $x_0 in (1,2)$, and Kantorovich’s theorem guarantees convergence of Newton’s method to $x_sol in (1,2)$

share|cite|improve this answer

answered Jul 30 at 11:16

Suhan Shetty

835

The solution given using Lambert’s function is very elegant. However, if you are interested in obtaining the solution using numerical methods, you can follow this:

Let $f(x) = 8^x-16x$. This a smooth function.

It's obvious from the graph of $f(x)$ that it cuts x-axis, and hence you are guaranteed a solution. In fact, you can see it has two solutions.

Otherwise, use IVT to verify this:
$f(1) = -6$ and $f(0) = 1$ and $f(1) = -6$ and $f(2) > 0$

By IVT, you are guaranteed a point $x_sol in (0,1)$ and one more in $(1,2)$ such that $f(x_sol)=0$. Since the function is monotonic ouside $(0,1)$ you will not get anymore solution points. )

You can use Newton's method to find the solution $x_sol$ in (0,1) as follows:

Choose an initial point $x_0$ for Newton's method in $(0,1)$.

$df(x) = 8^x ln 8-16$.

$|df(a)-df(b)| leq ln 8 |8^a-8^b|$ $forall$ $a,b in (0,1)$,

$implies |df(a)-df(b)| leq ln 8 |a-b|$

Thus, $df(x)$ is Lipschitz in $(0,1)$ with a Lipschitz ratio $M = ln 8$ in $(0,1)$,

Let $x_0 in (0,1)$

You can show that, $|f(x_0)M df^-1(x_0)^2| leq 0.5$

and hence you are guaranteed to converge to a solution $x_sol in (0,1)$ by Katorovich's theorem.

If you want the other solution point in $(1,2)$, choose $x_0 in (1,2)$, and Kantorovich’s theorem guarantees convergence of Newton’s method to $x_sol in (1,2)$

share|cite|improve this answer

answered Jul 30 at 11:16

Suhan Shetty

835

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 11:16

Suhan Shetty

835

answered Jul 30 at 11:16

Suhan Shetty

835

answered Jul 30 at 11:16

Suhan Shetty

835

835

add a comment | 

add a comment |