MHB Find k Such That Log Equation Has Real Solution

[anemone]

Find the values of k such that $\log (kx+1)-\log (x-k)=\log (2-x)$ has a unique real solution.

 

[Saitama]

Find the values of k such that $\log (kx+1)-\log (x-k)=\log (2-x)$ has a unique real solution.

Spoiler

The given equation can be rearranged as:

$$\frac{kx+1}{x-k}=2-x \Rightarrow x^2-2x+2k+1=0$$

For the above quadratic to have only one solution, the discriminant is zero i.e

$$4-4(2k+1)=0$$
$$\Rightarrow k=0$$

 

[anemone]

Spoiler

The given equation can be rearranged as:

$$\frac{kx+1}{x-k}=2-x \Rightarrow x^2-2x+2k+1=0$$

For the above quadratic to have only one solution, the discriminant is zero i.e

$$4-4(2k+1)=0$$
$$\Rightarrow k=0$$

Thank you for participating, Pranav!

But...what you have found is only one correct value of $k$ as there are many other values of $k$ which fit the case where the given equation has only a unique real solution...

 

[MarkFL]

Thank you for participating, Pranav!

But...what you have found is only one correct value of $k$ as there are many other values of $k$ which fit the case where the given equation has only a unique real solution...

What anemone means is find the values of $k$ for which the given equation has only 1 real solution. The use of the word unique implies there is only 1 solution. This is what has thrown us off. :D

 

[I like Serena]

Thank you for participating, Pranav!

But...what you have found is only one correct value of $k$ as there are many other values of $k$ which fit the case where the given equation has only a unique real solution...

Yep. Found another one! ;)
For $k=-2$ the equation would yield x=-1 or x=3, but x=3 is not allowed due to the restricted domain of the $\log$. However, x=-1 is allowed.
I seem to recall a couple of threads related to the domain of $\log$ a long long time ago.

 

[Saitama]

domain of $\log$ a long long time ago.

Hehe, yes, I remember. :P

I wait for the others to post their solution.

I am still trying to see where did I go wrong in the previous attempt. :)

 

[anemone]

Yep. Found another one! ;)
For $k=-2$ the equation would yield x=-1 or x=3, but x=3 is not allowed due to the restricted domain of the $\log$. However, x=-1 is allowed.

Yep! That is another right one! ;)

I seem to recall a couple of threads related to the domain of $\log$ a long long time ago.

You know, considering I have already posted quite many challenging problems here, I sometimes have to check before posting for another challenge problem and see if that is the one that has already been posted before and this is a very tiring checking process for me...for I need to go through many pages and scan through all threads of mine. This sounds very preposterous because I should have already thought to name the title of the threads cleverly (and clearly) for future reference...I am wondering from time to time if I can rename the titles of all of my threads...

(Edit: I just found another reliable and fast method to ascertain whether the problem at hand has already been posted before or not!)(Star):D

Hehe, yes, I remember. :P

I wait for the others to post their solution.

I am still trying to see where did I go wrong in the previous attempt. :)

Thanks...and I encourage you to keep trying!(Tongueout)

 

[Saitama]

Is the answer $(-\infty,-1/2] \cup{0}$? I guess this wrong because I had to do a lot of guesswork to reach this answer.

 

[anemone]

Is the answer $(-\infty,-1/2] \cup{0}$? I guess this wrong because I had to do a lot of guesswork to reach this answer.

Nope, that is not the answer to it.

I'm so so sorry, Pranav...

Edit: Perhaps you want to think along the line about the domain of logarithm functions...:)

 

[Saitama]

Nope, that is not the answer to it.

I'm so so sorry, Pranav...

Yes, the answer is incorrect, I found the error in my working.

I will quietly watch this thread from now. :)

 

[Opalg]

Find the values of k such that $\log (kx+1)-\log (x-k)=\log (2-x)$ has a unique real solution.

[sp]To avoid any danger of dividing by $0$, write the equation as $\log (kx+1) = \log (x-k) + \log (2-x)$. Now exponentiate: $kx+1 = (x-k)(2-x)$. That simplifies to $x^2 - 2x + 1 = 0$, which is independent of $k$ and has the unique solution $x=1.$

So where does $k$ come into it? The only remaining thing to do is to check that the logarithms are defined when $x=1$. For that, we need $k+1>0$, $1-k>0$ and $2-1>0$. The last of those is obviously no problem. The other two conditions require that $\boxed{|k|<1}$.[/sp]

 

[Saitama]

[sp]To avoid any danger of dividing by $0$, write the equation as $\log (kx+1) = \log (x-k) + \log (2-x)$. Now exponentiate: $kx+1 = (x-k)(2-x)$. That simplifies to $x^2 - 2x + 1 = 0$, which is independent of $k$ and has the unique solution $x=1.$

So where does $k$ come into it? The only remaining thing to do is to check that the logarithms are defined when $x=1$. For that, we need $k+1>0$, $1-k>0$ and $2-1>0$. The last of those is obviously no problem. The other two conditions require that $\boxed{|k|<1}$.[/sp]

I still haven't reached an answer but k=0.2 doesn't satisfy the conditions. For this value of k, there are complex roots.

The equation simplifies to $(x-1)^2=-2k$, it is easy to see that k must be less than or equal to zero for the equation to have roots.

 

[kaliprasad]

Spoiler

The given equation can be rearranged as:

$$\frac{kx+1}{x-k}=2-x \Rightarrow x^2-2x+2k+1=0$$

For the above quadratic to have only one solution, the discriminant is zero i.e

$$4-4(2k+1)=0$$
$$\Rightarrow k=0$$

I start from Pranav’s 1st solution

Now I proceed with the other part

Spoiler

Or

F(x) = x^2−2x+2k+1=0
Has one solution x < 2 as log (2-x) is undefined for x > 2

f(x) is + infinite at + infinite and – infinite so for a solution to exist f(2) < 0 or

2k + 1 < 0 => k < - ½

Now as log(kx + 1) is taken (kx + 1) > 0 or x < - 1/k ( - sign is taken as k is –ve)

Now solution

= (2 – sqrt( 4 - 4k – 4))/2 < - 1/k

Or ( 1- sqrt(-k)) < - 1/k

Because k is –ve and dividing or multiplying by k switches > to < vice versa put –k = p

So ( 1- sqrt(p)) < 1/p

Or (1- 1/p) < sqrt(p)

Solution is becoming complex and I dare not proceed but I would let some one to proceed from here.
I might have done a mistake

 

[anemone]

Is the answer $(-\infty,-1/2] \cup{0}$? I guess this wrong because I had to do a lot of guesswork to reach this answer.

Shame on me, Pranav! It appears that we have found the same answer, which I'm not completely sure is correct...at first I solved it and got the solution set for $k$ to be $[-\dfrac{1}{2},0]$. And I was so happy, but I didn't check to see if it's correct or not!

Now, after reading posts from Opalg and kaliprasad, I checked my work again, and I found the same answer as you. But, I must also tell you I am not entirely sure if ${k: -\infty<k \le -\dfrac{1}{2}}$ is the answer because to me, Opalg is always right.

I apologize for that.

[sp]To avoid any danger of dividing by $0$, write the equation as $\log (kx+1) = \log (x-k) + \log (2-x)$. Now exponentiate: $kx+1 = (x-k)(2-x)$. That simplifies to $x^2 - 2x + 1 = 0$, which is independent of $k$ and has the unique solution $x=1.$

So where does $k$ come into it? The only remaining thing to do is to check that the logarithms are defined when $x=1$. For that, we need $k+1>0$, $1-k>0$ and $2-1>0$. The last of those is obviously no problem. The other two conditions require that $\boxed{|k|<1}$.[/sp]

Hi Opalg,

Thanks for participating!:)

But Opalg, my answer is different than yours. Could you please check my work?

My solution:

First off, it's very important and crucial to always find the domain of the original given logarithmic equation and in this problem, we have

[TABLE="class: grid, width: 600"]
[TR]
[TD]$kx+1>0\;\;\;$[/TD]
[TD]$2-x>0\;\;\;$

$-x>-2$

$x<2$[/TD]
[TD]$x-k>0\;\;\;$

$x>k$[/TD]
[TD]That is, the solution sets that we have for $x$ and $k$ must satisfy these three inequalities simultaneously.[/TD]
[/TR]
[/TABLE]

Also, the original equation $\log (kx+1)-\log (x-k)=\log (2-x) \rightarrow \log (kx+1)=\log ((2-x)(x-k))$ is equivalent to the quadratic equation $kx+1=(2-x)(x-k) \rightarrow x^2-2x+2k+1=0$ after we equate the arguments of the log functions.

The quadratic equation $x^2-2x+2k+1=0$ has the two solutions, namely $x=1+\sqrt{-2k}$ and $x=1-\sqrt{-2k}$.

At this point, we can conclude that

1. If $k=0$, we achieved what we wanted and $k=0$ is a solution to the problem,

2. $k$ cannot be greater than zero,

2. We need to think also the case when $k<0$.

Bearing in mind the restriction given by the problem is the given equation has a unique real solution, we thus consider

[TABLE="class: grid, width: 600"]
[TR]
[TD]$k<0$, $x>k$ and $x=1-\sqrt{-2k}$[/TD]
[TD]and[/TD]
[TD]$k<0$, $x<2$ and $x=1+\sqrt{-2k}$[/TD]
[/TR]
[/TABLE]

[TABLE="class: grid, width: 500"]
[TR]
[TD]If $1-\sqrt{-2k}>k$ is true,[/TD]
[TD]then[/TD]
[TD]$1+\sqrt{-2k}\ge 2$ must hold.[/TD]
[/TR]
[/TABLE]

If we play with the inequality below a bit, we see that

$1-\sqrt{-2k}>k$

$1-k>\sqrt{-2k}$

$(1-k)^2>(\sqrt{-2k})^2$

$1-2k+k^2>-2k$

$1+k^2>0$

That is, $1-\sqrt{-2k}>k$ is always true for all $k \in R$.

In this case, in order to find the solution set for $k$, which is the answer to the problem, we must set

$1+\sqrt{-2k} \ge 2$

$\sqrt{-2k} \ge 1$

$-2k \ge 1$

$k \le -\frac{1}{2}$

Hence, the solution set for $k$ is ${k:k \le -\frac{1}{2}, k=0}$.

- - - Updated - - -

I start from Pranav’s 1st solution

Now I proceed with the other part

Spoiler

Or

F(x) = x^2−2x+2k+1=0
Has one solution x < 2 as log (2-x) is undefined for x > 2

f(x) is + infinite at + infinite and – infinite so for a solution to exist f(2) < 0 or

2k + 1 < 0 => k < - ½

Now as log(kx + 1) is taken (kx + 1) > 0 or x < - 1/k ( - sign is taken as k is –ve)

Now solution

= (2 – sqrt( 4 - 4k – 4))/2 < - 1/k

Or ( 1- sqrt(-k)) < - 1/k

Because k is –ve and dividing or multiplying by k switches > to < vice versa put –k = p

So ( 1- sqrt(p)) < 1/p

Or (1- 1/p) < sqrt(p)

Solution is becoming complex and I dare not proceed but I would let some one to proceed from here.
I might have done a mistake

Hey kaliprasad, yes, if we approach it like you did, things can become excessively complicated...

BTW, thanks for participating!

 

[kaliprasad]

Shame on me, Pranav! It appears that we have found the same answer, which I'm not completely sure is correct...at first I solved it and got the solution set for $k$ to be $[-\dfrac{1}{2},0]$. And I was so happy, but I didn't check to see if it's correct or not!

Now, after reading posts from Opalg and kaliprasad, I checked my work again, and I found the same answer as you. But, I must also tell you I am not entirely sure if ${k: -\infty<k \le -\dfrac{1}{2}}$ is the answer because to me, Opalg is always right.

I apologize for that.



Hi Opalg,

Thanks for participating!:)

But Opalg, my answer is different than yours. Could you please check my work?

My solution:

First off, it's very important and crucial to always find the domain of the original given logarithmic equation and in this problem, we have

[TABLE="class: grid, width: 600"]
[TR]
[TD]$kx+1>0\;\;\;$[/TD]
[TD]$2-x>0\;\;\;$

$-x>-2$

$x<2$[/TD]
[TD]$x-k>0\;\;\;$

$x>k$[/TD]
[TD]That is, the solution sets that we have for $x$ and $k$ must satisfy these three inequalities simultaneously.[/TD]
[/TR]
[/TABLE]

Also, the original equation $\log (kx+1)-\log (x-k)=\log (2-x) \rightarrow \log (kx+1)=\log ((2-x)(x-k))$ is equivalent to the quadratic equation $kx+1=(2-x)(x-k) \rightarrow x^2-2x+2k+1=0$ after we equate the arguments of the log functions.

The quadratic equation $x^2-2x+2k+1=0$ has the two solutions, namely $x=1+\sqrt{-2k}$ and $x=1-\sqrt{-2k}$.

At this point, we can conclude that

1. If $k=0$, we achieved what we wanted and $k=0$ is a solution to the problem,

2. $k$ cannot be greater than zero,

2. We need to think also the case when $k<0$.

Bearing in mind the restriction given by the problem is the given equation has a unique real solution, we thus consider

[TABLE="class: grid, width: 600"]
[TR]
[TD]$k<0$, $x>k$ and $x=1-\sqrt{-2k}$
[/TD]
[TD]and[/TD]
[TD]$k<0$, $x<2$ and $x=1+\sqrt{-2k}$[/TD]
[/TR]
[/TABLE]

[TABLE="class: grid, width: 500"]
[TR]
[TD]If $1-\sqrt{-2k}>k$ is true,[/TD]
[TD]then[/TD]
[TD]$1+\sqrt{-2k}>2$ must hold.[/TD]
[/TR]
[/TABLE]

If we play with the inequality below a bit, we see that

$1-\sqrt{-2k}>k$

$1-k>\sqrt{-2k}$

$(1-k)^2>(\sqrt{-2k})^2$

$1-2k+k^2>-2k$

$1+k^2>0$

That is, $1-\sqrt{-2k}>k$ is always true for all $k \in R$.

In this case, in order to find the solution set for $k$, which is the answer to the problem, we must set

$1+\sqrt{-2k}>2$

$\sqrt{-2k}>1$

$-2k>1$

$k<-\frac{1}{2}$

Hence, the solution set for $k$ is ${k:k<-\frac{1}{2}, k=0}$.

- - - Updated - - -
Hey kaliprasad, yes, if we approach it like you did, things can become excessively complicated...

BTW, thanks for participating!

Hello Anemone,
I appreciate your statement. Bout none of the above ans imply that kx +1 >0 ( I am sorry if it is implicit some where which I did not understand) and I formulated the equation to put a limit on x

 

[anemone]

Hello Anemone,
I appreciate your statement. Bout none of the above ans imply that kx +1 >0 ( I am sorry if it is implicit some where which I did not understand) and I formulated the equation to put a limit on x

I'm sorry kaliprasad for not covering the explanation in my previous post how the solution set of $k$ implies too that $kx+1>0$ must also be true.

Let me add the explanation in this post:

We know we now have only one real solution to the problem, namely $x=1-\sqrt{-2k}$.

The solution suggests that $k<-\dfrac{1}{2}$, $\therefore -2k>1$ which then tells us $x=1-\sqrt{-2k}<0$.

Since $k<-\dfrac{1}{2}$ and $x<0$, then $kx+1$ is always greater than zero. Agree

 

[Saitama]

I'm sorry kaliprasad for not covering the explanation in my previous post how the solution set of $k$ implies too that $kx+1>0$ must also be true.

Let me add the explanation in this post:

We know we now have only one real solution to the problem, namely $x=1-\sqrt{-2k}$.

The solution suggests that $k<-\dfrac{1}{2}$, $\therefore -2k>1$ which then tells us $x=1-\sqrt{-2k}<0$.

Since $k<-\dfrac{1}{2}$ and $x<0$, then $kx+1$ is always greater than zero. Agree

Sorry if this is going to be dumb but Wolfram Alpha doesn't give any solutions to:
$$1-\sqrt{-2k}>-\frac{1}{k}$$

1-sqrt(-2k)>-1/k - Wolfram|Alpha

EDIT: That was actually very dumb. I should have fed Wolfram Alpha with this:

k(1-sqrt(-2k))+1>0 - Wolfram|Alpha

 

[anemone]

Sorry if this is going to be dumb but Wolfram Alpha doesn't give any solutions to:
$$1-\sqrt{-2k}>-\frac{1}{k}$$

1-sqrt(-2k)>-1/k - Wolfram|Alpha

EDIT: That was actually very dumb. I should have fed Wolfram Alpha with this:

k(1-sqrt(-2k))+1>0 - Wolfram|Alpha

So, everything seems okay now?

 

[Saitama]

So, everything seems okay now?

I guess everything's ok. So then the final answer is $(-\infty,-1/2] \cup {0} $?

 

[anemone]

I guess everything's ok. So then the final answer is $(-\infty,-1/2] \cup {0} $?

Hmm...I think so.

I'm sorry though, because I don't have the answers for the vast majority of the challenge problems that I posted here for reference.

 

[MarkFL]

There seems to be some confusion in this thread, so I thought to see if I could add to it. (Heidy)

I think we can all agree that when $k=0$ this leads to the case where there is one real solution only, $x=1$. Otherwise we obtain the roots:

$$x=1\pm\sqrt{-2k}$$

In order for either of these roots to be real, we require $k<0$. (I use a strict inequality because we have already accounted for $k=0$

Analyzing the arguments of the log functions, we find:

a) $$x<2$$

Using the larger root, and making it greater than or equal to 2, and the smaller root less than 2, thereby ensuring only 1 root is real, we obtain

i) $$1+\sqrt{-2k}\ge2$$

$$-2k\ge1$$

$$k\le-\frac{1}{2}$$

ii) $$1-\sqrt{-2k}<2$$

$$-\sqrt{-2k}<1$$

$$-2k\ge0$$

$$k\le0$$

We need both of these to be true, hence:

$$k\le-\frac{1}{2}$$

b) $$x>k$$

We will find here that both roots must be larger than $k$, otherwise there are no solutions for $k$. Setting the smaller root greater than $k$:

$$1-\sqrt{-2k}>k$$

$$k\le0$$

c) $$kx>-1$$

In order for solutions to exist, we need:

i) $$k\left(1+\sqrt{-2k} \right)\le-1$$

$$k\le-\frac{1}{2}$$

ii) $$k\left(1-\sqrt{-2k} \right)>-1$$

$$k\le0$$

We need both of these to be true, hence:

$$k\le-\frac{1}{2}$$

Thus, we find the solutions:

$$k=0,\,k\in\,\left(-\infty,-\frac{1}{2} \right]$$

 

[Opalg]

I guess everything's ok. So then the final answer is $(-\infty,-1/2] \cup {0} $?

My previous attempt at an answer obviously went off the rails right from the start. I agree with the answer $(-\infty,-1/2] \cup {0} $.

 

[kaliprasad]

I was right till the point

Spoiler

Or

F(x) = x^2−2x+2k+1=0
Has one solution x < 2 as log (2-x) is undefined for x > 2

f(x) is + infinite at + infinite and – infinite so for a solution to exist f(2) < 0 or

2k + 1 < 0 => k < - ½

Then to check that kx + 1 > 0 I complicated the things

I should have realized that

if k < -1/2
x = 1- sqrt(-2k) < 0

as kx + 1 > 0 is satisfied.

Thanks to Anemone for clarifying the things and solution is not complex

 

[I like Serena]

I see some confusion whether the solution $k=-\frac 1 2$ is included or not.
Can we include it please?

That is since $kx + 1 > 0$ is not satisfied for $k=-\frac 1 2$ and $x=2$, so we have only $x=0$ as solution, as required.

 

[MarkFL]

I see some confusion whether the solution $k=-\frac 1 2$ is included or not.
Can we include it please?

That is since $kx + 1 > 0$ is not satisfied for $k=-\frac 1 2$ and $x=2$, so we have only $x=0$ as solution, as required.

Yes, I see what I did wrong. (Doh)

I fixed my post above...thanks for catching this. :D

 

[anemone]

I see some confusion whether the solution $k=-\frac 1 2$ is included or not.
Can we include it please?

That is since $kx + 1 > 0$ is not satisfied for $k=-\frac 1 2$ and $x=2$, so we have only $x=0$ as solution, as required.

Oops...I also just fixed my solution post, thanks I like Serena!