Mixing
Problem with logarithmic equation. [closed]
I want to solve this problem, when I tried to differentiate both sides I didn't get the answer. I need to know why, and, secondly I need to know how to solve this problem:
$$ln x = frac{1}{x}-1$$
Differentiating both sides with respect to $x$ does not work.
ADDED: I need a solution method that does not use graphing, sketching, or "trying" (also called "guess-and-check"). The method needs to find all the solutions.
calculus algebra-precalculus
edited Nov 20 '18 at 1:06
Rory Daulton
29.3k53254
asked Nov 19 '18 at 21:49
mahmoud solyman
4
closed as unclear what you're asking by Chase Ryan Taylor, Lord Shark the Unknown, Leucippus, jgon, Cesareo Nov 20 '18 at 1:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 2 more comments
I want to solve this problem, when I tried to differentiate both sides I didn't get the answer. I need to know why, and, secondly I need to know how to solve this problem:
$$ln x = frac{1}{x}-1$$
Differentiating both sides with respect to $x$ does not work.
ADDED: I need a solution method that does not use graphing, sketching, or "trying" (also called "guess-and-check"). The method needs to find all the solutions.
calculus algebra-precalculus
edited Nov 20 '18 at 1:06
Rory Daulton
29.3k53254
asked Nov 19 '18 at 21:49
mahmoud solyman
4
closed as unclear what you're asking by Chase Ryan Taylor, Lord Shark the Unknown, Leucippus, jgon, Cesareo Nov 20 '18 at 1:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
6
What's the question? Do you want to find a particular value of $x$ such that $ln(x) =frac 1x-1$? Why would you differentiate both sides?
– lulu
Nov 19 '18 at 21:50
Hint: it is easy to guess a solution but then you need to argue that the solution is unique. But a quick sketch of the two functions should tell you how to do that.
– lulu
Nov 19 '18 at 21:53
Yes i want the particular solution and want to know if it is the only solution without graphing
– mahmoud solyman
Nov 19 '18 at 21:55
If you differentiate both sides you will get that the left hand side is an increasing function and the right hand side is a decreasing function. Which would suggest that there is at most one solution. Though, It would be cleaner to say that $frac {d}{dx} [ln x - (frac 1x - 1)] > 0$
– Doug M
Nov 19 '18 at 21:55
It's clear thank you , and how can i solve the equation without sketching 😅
– mahmoud solyman
Nov 19 '18 at 22:00
|
show 2 more comments
I want to solve this problem, when I tried to differentiate both sides I didn't get the answer. I need to know why, and, secondly I need to know how to solve this problem:
$$ln x = frac{1}{x}-1$$
Differentiating both sides with respect to $x$ does not work.
ADDED: I need a solution method that does not use graphing, sketching, or "trying" (also called "guess-and-check"). The method needs to find all the solutions.
calculus algebra-precalculus
edited Nov 20 '18 at 1:06
Rory Daulton
29.3k53254
asked Nov 19 '18 at 21:49
mahmoud solyman
4
I want to solve this problem, when I tried to differentiate both sides I didn't get the answer. I need to know why, and, secondly I need to know how to solve this problem:
$$ln x = frac{1}{x}-1$$
Differentiating both sides with respect to $x$ does not work.
ADDED: I need a solution method that does not use graphing, sketching, or "trying" (also called "guess-and-check"). The method needs to find all the solutions.
calculus algebra-precalculus
calculus algebra-precalculus
edited Nov 20 '18 at 1:06
Rory Daulton
29.3k53254
asked Nov 19 '18 at 21:49
mahmoud solyman
4
edited Nov 20 '18 at 1:06
Rory Daulton
29.3k53254
asked Nov 19 '18 at 21:49
mahmoud solyman
4
edited Nov 20 '18 at 1:06
Rory Daulton
29.3k53254
edited Nov 20 '18 at 1:06
Rory Daulton
29.3k53254
edited Nov 20 '18 at 1:06
Rory Daulton
29.3k53254
29.3k53254
asked Nov 19 '18 at 21:49
mahmoud solyman
4
asked Nov 19 '18 at 21:49
mahmoud solyman
4
asked Nov 19 '18 at 21:49
mahmoud solyman
4
4
closed as unclear what you're asking by Chase Ryan Taylor, Lord Shark the Unknown, Leucippus, jgon, Cesareo Nov 20 '18 at 1:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Chase Ryan Taylor, Lord Shark the Unknown, Leucippus, jgon, Cesareo Nov 20 '18 at 1:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
6
What's the question? Do you want to find a particular value of $x$ such that $ln(x) =frac 1x-1$? Why would you differentiate both sides?
– lulu
Nov 19 '18 at 21:50
Hint: it is easy to guess a solution but then you need to argue that the solution is unique. But a quick sketch of the two functions should tell you how to do that.
– lulu
Nov 19 '18 at 21:53
Yes i want the particular solution and want to know if it is the only solution without graphing
– mahmoud solyman
Nov 19 '18 at 21:55
If you differentiate both sides you will get that the left hand side is an increasing function and the right hand side is a decreasing function. Which would suggest that there is at most one solution. Though, It would be cleaner to say that $frac {d}{dx} [ln x - (frac 1x - 1)] > 0$
– Doug M
Nov 19 '18 at 21:55
It's clear thank you , and how can i solve the equation without sketching 😅
– mahmoud solyman
Nov 19 '18 at 22:00
|
show 2 more comments
6
What's the question? Do you want to find a particular value of $x$ such that $ln(x) =frac 1x-1$? Why would you differentiate both sides?
– lulu
Nov 19 '18 at 21:50
Hint: it is easy to guess a solution but then you need to argue that the solution is unique. But a quick sketch of the two functions should tell you how to do that.
– lulu
Nov 19 '18 at 21:53
Yes i want the particular solution and want to know if it is the only solution without graphing
– mahmoud solyman
Nov 19 '18 at 21:55
If you differentiate both sides you will get that the left hand side is an increasing function and the right hand side is a decreasing function. Which would suggest that there is at most one solution. Though, It would be cleaner to say that $frac {d}{dx} [ln x - (frac 1x - 1)] > 0$
– Doug M
Nov 19 '18 at 21:55
It's clear thank you , and how can i solve the equation without sketching 😅
– mahmoud solyman
Nov 19 '18 at 22:00
6
6
What's the question? Do you want to find a particular value of $x$ such that $ln(x) =frac 1x-1$? Why would you differentiate both sides?
– lulu
Nov 19 '18 at 21:50
What's the question? Do you want to find a particular value of $x$ such that $ln(x) =frac 1x-1$? Why would you differentiate both sides?
– lulu
Nov 19 '18 at 21:50
Hint: it is easy to guess a solution but then you need to argue that the solution is unique. But a quick sketch of the two functions should tell you how to do that.
– lulu
Nov 19 '18 at 21:53
Hint: it is easy to guess a solution but then you need to argue that the solution is unique. But a quick sketch of the two functions should tell you how to do that.
– lulu
Nov 19 '18 at 21:53
Yes i want the particular solution and want to know if it is the only solution without graphing
– mahmoud solyman
Nov 19 '18 at 21:55
Yes i want the particular solution and want to know if it is the only solution without graphing
– mahmoud solyman
Nov 19 '18 at 21:55
If you differentiate both sides you will get that the left hand side is an increasing function and the right hand side is a decreasing function. Which would suggest that there is at most one solution. Though, It would be cleaner to say that $frac {d}{dx} [ln x - (frac 1x - 1)] > 0$
– Doug M
Nov 19 '18 at 21:55
If you differentiate both sides you will get that the left hand side is an increasing function and the right hand side is a decreasing function. Which would suggest that there is at most one solution. Though, It would be cleaner to say that $frac {d}{dx} [ln x - (frac 1x - 1)] > 0$
– Doug M
Nov 19 '18 at 21:55
It's clear thank you , and how can i solve the equation without sketching 😅
– mahmoud solyman
Nov 19 '18 at 22:00
It's clear thank you , and how can i solve the equation without sketching 😅
– mahmoud solyman
Nov 19 '18 at 22:00
|
show 2 more comments
1 Answer
1
active
oldest
votes
You say that you need a solution method without sketching and without "trying" (also called "guess-and-check"). Here is a solution using the Lambert W function. I assume that you want to find all real-number solutions to your equation.
$$ln x=frac 1x-1$$
$$e^{ln x}=e^{1/x-1}$$
$$x=e^{1/x}cdot e^{-1}$$
$$e=frac 1xcdot e^{1/x}$$
$$frac 1x=W(e)$$
That last transition is due to the definition of the Lambert W function. Continuing,
$$x=dfrac 1{W(e)}$$
$$x=frac 11$$
That last is a well-known special value of the Lambert W function. This is not "trying" or "guess-and-check", it is a specific value of a specific well-known function. So we end up with
$$x=1$$
We substitute that back into the original solution as a check, and it works.
This is an algebraic solution, using a well-known special function. There is one possible criticism of this solution method. The Lambert W function has infinitely many branches, two of them dealing with real number values. However, only one branch gives a real value at the parameter value $e$, and that real value is one. That justifies the use of only one branch of the function and one result. If you allow complex number solutions, other branches could be used to get the other values. (There are a few complications for complex number solutions of this equation that I'll ignore here--they are not relevant for real number solutions.) If you would like some of those other values, let me know and I can calculate close approximations of them for you.
answered Nov 20 '18 at 0:49
Rory Daulton
29.3k53254
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You say that you need a solution method without sketching and without "trying" (also called "guess-and-check"). Here is a solution using the Lambert W function. I assume that you want to find all real-number solutions to your equation.
$$ln x=frac 1x-1$$
$$e^{ln x}=e^{1/x-1}$$
$$x=e^{1/x}cdot e^{-1}$$
$$e=frac 1xcdot e^{1/x}$$
$$frac 1x=W(e)$$
That last transition is due to the definition of the Lambert W function. Continuing,
$$x=dfrac 1{W(e)}$$
$$x=frac 11$$
That last is a well-known special value of the Lambert W function. This is not "trying" or "guess-and-check", it is a specific value of a specific well-known function. So we end up with
$$x=1$$
We substitute that back into the original solution as a check, and it works.
This is an algebraic solution, using a well-known special function. There is one possible criticism of this solution method. The Lambert W function has infinitely many branches, two of them dealing with real number values. However, only one branch gives a real value at the parameter value $e$, and that real value is one. That justifies the use of only one branch of the function and one result. If you allow complex number solutions, other branches could be used to get the other values. (There are a few complications for complex number solutions of this equation that I'll ignore here--they are not relevant for real number solutions.) If you would like some of those other values, let me know and I can calculate close approximations of them for you.
answered Nov 20 '18 at 0:49
Rory Daulton
29.3k53254
add a comment |
You say that you need a solution method without sketching and without "trying" (also called "guess-and-check"). Here is a solution using the Lambert W function. I assume that you want to find all real-number solutions to your equation.
$$ln x=frac 1x-1$$
$$e^{ln x}=e^{1/x-1}$$
$$x=e^{1/x}cdot e^{-1}$$
$$e=frac 1xcdot e^{1/x}$$
$$frac 1x=W(e)$$
That last transition is due to the definition of the Lambert W function. Continuing,
$$x=dfrac 1{W(e)}$$
$$x=frac 11$$
That last is a well-known special value of the Lambert W function. This is not "trying" or "guess-and-check", it is a specific value of a specific well-known function. So we end up with
$$x=1$$
We substitute that back into the original solution as a check, and it works.
This is an algebraic solution, using a well-known special function. There is one possible criticism of this solution method. The Lambert W function has infinitely many branches, two of them dealing with real number values. However, only one branch gives a real value at the parameter value $e$, and that real value is one. That justifies the use of only one branch of the function and one result. If you allow complex number solutions, other branches could be used to get the other values. (There are a few complications for complex number solutions of this equation that I'll ignore here--they are not relevant for real number solutions.) If you would like some of those other values, let me know and I can calculate close approximations of them for you.
answered Nov 20 '18 at 0:49
Rory Daulton
29.3k53254
add a comment |
You say that you need a solution method without sketching and without "trying" (also called "guess-and-check"). Here is a solution using the Lambert W function. I assume that you want to find all real-number solutions to your equation.
$$ln x=frac 1x-1$$
$$e^{ln x}=e^{1/x-1}$$
$$x=e^{1/x}cdot e^{-1}$$
$$e=frac 1xcdot e^{1/x}$$
$$frac 1x=W(e)$$
That last transition is due to the definition of the Lambert W function. Continuing,
$$x=dfrac 1{W(e)}$$
$$x=frac 11$$
That last is a well-known special value of the Lambert W function. This is not "trying" or "guess-and-check", it is a specific value of a specific well-known function. So we end up with
$$x=1$$
We substitute that back into the original solution as a check, and it works.
This is an algebraic solution, using a well-known special function. There is one possible criticism of this solution method. The Lambert W function has infinitely many branches, two of them dealing with real number values. However, only one branch gives a real value at the parameter value $e$, and that real value is one. That justifies the use of only one branch of the function and one result. If you allow complex number solutions, other branches could be used to get the other values. (There are a few complications for complex number solutions of this equation that I'll ignore here--they are not relevant for real number solutions.) If you would like some of those other values, let me know and I can calculate close approximations of them for you.
answered Nov 20 '18 at 0:49
Rory Daulton
29.3k53254
You say that you need a solution method without sketching and without "trying" (also called "guess-and-check"). Here is a solution using the Lambert W function. I assume that you want to find all real-number solutions to your equation.
$$ln x=frac 1x-1$$
$$e^{ln x}=e^{1/x-1}$$
$$x=e^{1/x}cdot e^{-1}$$
$$e=frac 1xcdot e^{1/x}$$
$$frac 1x=W(e)$$
That last transition is due to the definition of the Lambert W function. Continuing,
$$x=dfrac 1{W(e)}$$
$$x=frac 11$$
That last is a well-known special value of the Lambert W function. This is not "trying" or "guess-and-check", it is a specific value of a specific well-known function. So we end up with
$$x=1$$
We substitute that back into the original solution as a check, and it works.
This is an algebraic solution, using a well-known special function. There is one possible criticism of this solution method. The Lambert W function has infinitely many branches, two of them dealing with real number values. However, only one branch gives a real value at the parameter value $e$, and that real value is one. That justifies the use of only one branch of the function and one result. If you allow complex number solutions, other branches could be used to get the other values. (There are a few complications for complex number solutions of this equation that I'll ignore here--they are not relevant for real number solutions.) If you would like some of those other values, let me know and I can calculate close approximations of them for you.
answered Nov 20 '18 at 0:49
Rory Daulton
29.3k53254
edited Nov 20 '18 at 11:09
answered Nov 20 '18 at 0:49
Rory Daulton
29.3k53254
answered Nov 20 '18 at 0:49
Rory Daulton
29.3k53254
answered Nov 20 '18 at 0:49
Rory Daulton
29.3k53254
29.3k53254
add a comment |
add a comment |
6
What's the question? Do you want to find a particular value of $x$ such that $ln(x) =frac 1x-1$? Why would you differentiate both sides?
– lulu
Nov 19 '18 at 21:50
Hint: it is easy to guess a solution but then you need to argue that the solution is unique. But a quick sketch of the two functions should tell you how to do that.
– lulu
Nov 19 '18 at 21:53
Yes i want the particular solution and want to know if it is the only solution without graphing
– mahmoud solyman
Nov 19 '18 at 21:55
If you differentiate both sides you will get that the left hand side is an increasing function and the right hand side is a decreasing function. Which would suggest that there is at most one solution. Though, It would be cleaner to say that $frac {d}{dx} [ln x - (frac 1x - 1)] > 0$
– Doug M
Nov 19 '18 at 21:55
It's clear thank you , and how can i solve the equation without sketching 😅
– mahmoud solyman
Nov 19 '18 at 22:00