Determining the form of the real solution set?

[diredragon]

Homework Statement

$\frac{log_{2^{x^2+2x+1}-1}(log_{2x^2 + 2x + 3}(x^2 - 2x)}{log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)} \geqslant 0$
The set of all real solutions to this inequality is of the form:
$(a)$ $(a,b) \cup (b,c)$, $(b)$ $(-\infty,a) \cup (c,\infty)$, $(c)$ $(a,b)$
for some real numbers $a,b,c$ such that $-\infty<a<b<c<+\infty$

Homework Equations

The Attempt at a Solution

First i found the points at which the numerator and the denominator equal $0$. I found that from the following:
$log_{2^{x^2+2x+1}-1}(log_{2x^2 + 2x + 3}(x^2 - 2x) = 0$
$log_{2x^2 + 2x + 3}(x^2 - 2x) = 1$
$2x^2 + 2x + 3 = x^2 - 2x$
from this quadratic equation i find that the zeros of the function are $x_1=-3$ and $x_2=-1$ and let them be $a=-1$ and $b=-3$ for now. i also found from the plot of the given quadratic equation that $(-\infty,-3)\cup(-1,\infty)$ makes function a positiv one and the values in between $(-3,-1)$ make it negative. I used that later for the rational number chart.
Similarly $log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10) = 0$
$x^2 + 6x + 10 = 1$
$x = -3$. This function is positive for every other value.
Now using the chart of rational numbers, determining when the whole ratio is positive and when is negative i get the positive values to range from
$(-\infty,-3) \cup (-1,\infty)$ which then makes my answer of the form $(-\infty,b) \cup (a, +\infty)$.
First of all, its not given, second, its wrong. The solution is i know the $(a)$ answer. I just don't know how to get to it. Could you help?

 

[RUber]

It looks like you have done well to eliminate the values which would drive a division by zero. However, there are other considerations you must apply to satisfy the domain for the logarithms.
First, your base may not be zero. $\log_0$ does not make any sense.
Second, if your base is positive, your input value may not be negative, $\forall a,b>0 \log_a (-b)$ does not exist.
However, I don't think that either of these is the cause of your trouble.

I would recommend rewriting this using some log rules.

Edit: incorrect.

Remove the outer log by putting both sides of the inequality in the exponent.

Edit: incorrect.

Then take another look.

 

[Mark44]

Homework Statement

$\frac{log_{2^{x^2+2x+1}-1}(log_{2x^2 + 2x + 3}(x^2 - 2x)}{log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)} \geqslant 0$
The set of all real solutions to this inequality is of the form:
$(a)$ $(a,b) \cup (b,c)$, $(b)$ $(-\infty,a) \cup (c,\infty)$, $(c)$ $(a,b)$
for some real numbers $a,b,c$ such that $-\infty<a<b<c<+\infty$

Homework Equations

3. The Attempt at a Solution [/B]
First i found the points at which the numerator and the denominator equal $0$. I found that from the following:
$log_{2^{x^2+2x+1}-1}(log_{2x^2 + 2x + 3}(x^2 - 2x) = 0$
$log_{2x^2 + 2x + 3}(x^2 - 2x) = 1$
$2x^2 + 2x + 3 = x^2 - 2x$
from this quadratic equation i find that the zeros of the function are $x_1=-3$ and $x_2=-1$ and let them be $a=-1$ and $b=-3$ for now. i also found from the plot of the given quadratic equation that $(-\infty,-3)\cup(-1,\infty)$ makes function a positiv one and the values in between $(-3,-1)$ make it negative. I used that later for the rational number chart.
Similarly $log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10) = 0$
$x^2 + 6x + 10 = 1$
$x = -3$. This function is positive for every other value.
Now using the chart of rational numbers, determining when the whole ratio is positive and when is negative i get the positive values to range from
$(-\infty,-3) \cup (-1,\infty)$ which then makes my answer of the form $(-\infty,b) \cup (a, +\infty)$.
First of all, its not given, second, its wrong. The solution is i know the $(a)$ answer. I just don't know how to get to it. Could you help?

Please don't overly decorate your text with various font and BBCode tags if you're not sure of what you're doing. In particular, don't mix BBCode bold tags ( [B] ) with LaTeX -- it screws up both.

 

[SammyS]

It looks like you have done well to eliminate the values which would drive a division by zero. However, there are other considerations you must apply to satisfy the domain for the logarithms.
First, your base may not be zero. $\log_0$ does not make any sense.
Second, if your base is positive, your input value may not be negative, $\forall a,b>0 \log_a (-b)$ does not exist.
However, I don't think that either of these is the cause of your trouble.

I would recommend rewriting this using some log rules.
$log_{2^{x^2 +2x+1} - 1 } ( log_{2x^2+2x+3} (x^2 - 2x ) - ( x^2 +6x + 10) ) \geq 0$
Remove the outer log by putting both sides of the inequality in the exponent of $(2^{x^2+2x+1} - 1)^y$
Then take another look.

 

[diredragon]

It looks like you have done well to eliminate the values which would drive a division by zero. However, there are other considerations you must apply to satisfy the domain for the logarithms.
First, your base may not be zero. $\log_0$ does not make any sense.
Second, if your base is positive, your input value may not be negative, $\forall a,b>0 \log_a (-b)$ does not exist.
However, I don't think that either of these is the cause of your trouble.

I would recommend rewriting this using some log rules.
$log_{2^{x^2 +2x+1} - 1 } ( log_{2x^2+2x+3} (x^2 - 2x ) - ( x^2 +6x + 10) ) \geq 0$
Remove the outer log by putting both sides of the inequality in the exponent of $(2^{x^2+2x+1} - 1)^y$
Then take another look.

So $(2^{x^2 + 2x + 1} - 1)^y = log_{2x^2 +2x +3 }(x^2 - 2x) - (x^2 + 6x + 10)$

 

[SammyS]

It looks like you have done well to eliminate the values which would drive a division by zero. However, there are other considerations you must apply to satisfy the domain for the logarithms.
First, your base may not be zero. $\log_0$ does not make any sense.
Second, if your base is positive, your input value may not be negative, $\forall a,b>0 \log_a (-b)$ does not exist.
However, I don't think that either of these is the cause of your trouble.

I would recommend rewriting this using some log rules.
$log_{2^{x^2 +2x+1} - 1 } ( log_{2x^2+2x+3} (x^2 - 2x ) - ( x^2 +6x + 10) ) \geq 0$
Remove the outer log by putting both sides of the inequality in the exponent of $(2^{x^2+2x+1} - 1)^y$
Then take another look.

That's not correct.

$\displaystyle \frac{log_{\,b}(A)}{log_{\,b}(C)}\neq log_{\,b}(A)-log_{\,b}(C)\$

However, the change of base formula may come in handy.

 

[diredragon]

That's not correct.

$\displaystyle \frac{log_{\,b}(A)}{log_{\,b}(C)}\neq log_{\,b}(A)-log_{\,b}(C)\$

However, the change of base formula may come in handy.

$\frac{log_{2^{x^2+2x+1}-1}(log_{2x^2 + 2x + 3}(x^2 - 2x)}{log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)}$
This was rewritten correctly i think
$log_{2^{x^2+2x+1}-1}((log_{2x^2 + 2x + 3}(x^2 - 2x) - (x^2 + 6x + 10))$

 

[RUber]

edit: incorrect.

 

[SammyS]

Putting both sides of the inequality into the exponent should give you something like:
$log_{2x^2 +2x +3 }(x^2 - 2x) - (x^2 + 6x + 10) \geq (2^{x^2 + 2x + 1} - 1)^0.$

Yes. It's worse than I thought. -- that is if you consider one incorrect form to be worse than another.

What RUber seems to be saying in post #2 is that $\displaystyle \ \frac{\log_{\,b}(A)}{\log_{\,b}(C)}= \log_{\,b}(A-C)\$, which is absolutely wrong.

Added in Edit:

What is the case, is that $\displaystyle \ \log_{\,b}\left(\frac{A}{C}\right)= \log_{\,b}(A)-\log_{\,b}(C)\$.

Also true is, $\displaystyle \ \frac{\log_{\,b}(A)}{\log_{\,b}(C)}= \log_{\,C}(A)\$

 

[RUber]

Wow-- thanks SammyS. I clearly was not paying attention.
So then using the change of base to remove the quotient, and then exponentiating to remove the logs would be one way to resolve this.
The other way would be to compare signs of the top and bottom -- but I think that simplifying would help.

 

[SammyS]

Wow-- thanks SammyS. I clearly was not paying attention.
So then using the change of base to remove the quotient, and then exponentiating to remove the logs would be one way to resolve this.
The other way would be to compare signs of the top and bottom -- but I think that simplifying would help.

RUber, you're right. Considerable simplification can be done. Also, as you point out, attention must be paid to several details.

The "outer' logarithm in numerator has the same base as the logarithm in the denominator, so a change of base is in order. That leaves us with the log of a log to compare with zero.

As you point out, the base of any of these logarithms must be positive and must not equal 1 . That does eliminate at least one x value and may give OP his value for b, the deleted point.

 

[diredragon]

RUber, you're right. Considerable simplification can be done. Also, as you point out, attention must be paid to several details.

The "outer' logarithm in numerator has the same base as the logarithm in the denominator, so a change of base is in order. That leaves us with the log of a log to compare with zero.

As you point out, the base of any of these logarithms must be positive and must not equal 1 . That does eliminate at least one x value and may give OP his value for b, the deleted point.

So you suggest rewritting to $\log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0$

 

[RUber]

That's right. And solve that like you did for the polynomial in your first post. Pay attention to the inequality. Finally, check for any values that must be excluded from the original form of the problem.

 

[diredragon]

That's right. And solve that like you did for the polynomial in your first post. Pay attention to the inequality. Finally, check for any values that must be excluded from the original form of the problem.

i went on like this $\log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0$, $\log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1$, $2x^2 + 2x + 3 \ge x^2 - 2x$, $x_1 = -3, x_2 = -1$, also from the change of base formula it is necessary that $x^2 + 6x + 10 \neq 1$ giving same results and that $2^{x^2 +2x + 1} - 1 \neq 1$ giving $x_3=0, x_4=-2$
what to do with these results?

 

[SammyS]

i went on like this $\log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0$, $\log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1$, $2x^2 + 2x + 3 \ge x^2 - 2x$, $x_1 = -3, x_2 = -1$, also from the change of base formula it is necessary that $x^2 + 6x + 10 \neq 1$ giving same results and that $2^{x^2 +2x + 1} - 1 \neq 1$ giving $x_3=0, x_4=-2$
what to do with these results?

What does $\displaystyle \ 2^{x^2 +2x + 1} - 1 = 1\$ imply regarding your initial expression (prior to doing the charge of base)?

 

[diredragon]

it implies $\log_{2x^2 + 2x + 3} (x^2 - 2x) = x^2 + 6x + 10 = 1$ the base is 1, so it must be 1.

 

[SammyS]

it implies $\log_{2x^2 + 2x + 3} (x^2 - 2x) = x^2 + 6x + 10 = 1$ the base is 1, so it must be 1.

But log₁ is not defined as a function.

 

[RUber]

i went on like this $\log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0$, $\log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1$, $2x^2 + 2x + 3 \ge x^2 - 2x$, $x_1 = -3, x_2 = -1$, also from the change of base formula it is necessary that $x^2 + 6x + 10 \neq 1$ giving same results and that $2^{x^2 +2x + 1} - 1 \neq 1$ giving $x_3=0, x_4=-2$
what to do with these results?

Was the appropriate range for the inequality $(x_1, x_2)$ or $( -\infty , x_1)\cup ( x_2, \infty)$?
Depending on which interval you choose, only one of $x_3, x_4$ would be excluded from the solution set.
Edit
It looks like you have the wrong inequality...
$\log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1$, and $2x^2 + 2x + 3 \ge x^2 - 2x$ are not consistent.
To remove the log, you are doing:
$(2x^2 + 2x + 3)^{\log_{2x^2 + 2x + 3} (x^2 - 2x)} \ge (2x^2 + 2x + 3)^1$
Which should give you:
$x^2 - 2x \ge 2x^2 + 2x + 3$.
This will give you the appropriate interval for the solution once you remove your excluded point.

 

[diredragon]

Was the appropriate range for the inequality $(x_1, x_2)$ or $( -\infty , x_1)\cup ( x_2, \infty)$?
Depending on which interval you choose, only one of $x_3, x_4$ would be excluded from the solution set.
Edit
It looks like you have the wrong inequality...
$\log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1$, and $2x^2 + 2x + 3 \ge x^2 - 2x$ are not consistent.
To remove the log, you are doing:
$(2x^2 + 2x + 3)^{\log_{2x^2 + 2x + 3} (x^2 - 2x)} \ge (2x^2 + 2x + 3)^1$
Which should give you:
$x^2 - 2x \ge 2x^2 + 2x + 3$.
This will give you the appropriate interval for the solution once you remove your excluded point.

you are right, i have made a wrong inequality. so $x^2 - 2x \ge 2x^2 + 2x + 3$ giving me $x_1 = -1, x_2=-3$. now since i get from the change of base formula that $x^2 - 2x > 1$ and i have what i have written above can i put the x-es i get to see which is excluded? i get by those means the $-1, -2, -3$ fit while $0$ should be excluded. I don't know if you can do it this way

 

[diredragon]

But log₁ is not defined as a function.

then it implies that the initial inequality involves undefined functions...what else could it be

 

[RUber]

you are right, i have made a wrong inequality. so $x^2 - 2x \ge 2x^2 + 2x + 3$ giving me $x_1 = -1, x_2=-3$. now since i get from the change of base formula that $x^2 - 2x > 1$ and i have what i have written above can i put the x-es i get to see which is excluded? i get by those means the $-1, -2, -3$ fit while $0$ should be excluded. I don't know if you can do it this way

I am not sure I understand what you mean.

From the inequality $x^2 - 2x \ge 2x^2 + 2x + 3$, you should get an interval which satisfies the condition, not just two x values.

From the condition that the base of a log cannot be 1 or 0, you get single excluded values (x1,x2,x3,x4), which you have found to be -3, -2, -1, 0. Two of those are the endpoints of your interval, and one of the other two is included in that interval.

 

[SammyS]

then it implies that the initial inequality involves undefined functions...what else could it be

Right!

It's undefined for what ever value of x makes $\displaystyle \ 2^{x^2 +2x + 1} - 1 = 1\,,\$ because then you have an expression with log₁ in it.

You really should look carefully at the possible values for the base of each of your logarithms, looking for any possibly problem values.

 

[diredragon]

Ok, so $0$ makes it undefined. Other vakues are ok i guess. So i would pick $(a,b) \cup (b,c)$ it has three values so only because of that i would choose it.

 

[SammyS]

Ok, so $0$ makes it undefined. Other values are ok i guess. So i would pick $(a,b) \cup (b,c)$ it has three values so only because of that i would choose it.

Yes, $\displaystyle \ 2^{x^2 +2x + 1} - 1\$ evaluated at x = 0 is 1, if that's what you mean. However, x = 0 should not be part of your potential solution set anyway (that is if you are following RUber).

Is there any other way to get 2¹ - 1 , or worse yet, 2⁰ - 1 ?So, what are your values for a, b, and c, and are you sure that there are no other issues there?

 

[diredragon]

I redid the problem again and i think i undestand it now. The solution is $(-3, -2) \cup (-2, -1)$ with a gap in $-2$ because the base there is $1$.
The expression $x^2 - 2x \ge 2x^2 + 2x + 3$ holds for numbers between $-3, -1$ any subtleties i might have missed?

 

[SammyS]

I redid the problem again and i think i undestand it now. The solution is $(-3, -2) \cup (-2, -1)$ with a gap in $-2$ because the base there is $1$.
The expression $x^2 - 2x \ge 2x^2 + 2x + 3$ holds for numbers between $-3, -1$ any subtleties i might have missed?

That appears to be the correct answer.

Are there any other subtleties?

Well, none that change the final answer, but there are some additional items to consider.

Some details:
$\displaystyle \ x^2 - 2x \ge 2x^2 + 2x + 3 \$ holds for x in the interval $\ \left[ -3,\, -1 \right] \$.

We've covered the reason for deleting $-2$ from that interval. Why are the end points also deleted?.

Well, previously mentioned is that $\displaystyle \ 2^{x^2 +2x + 1} - 1$ evaluates to 0 at $\ x=-1\$ and to 1 at $\ x=-2,\,0\$.

What's up with $\ -3\$? Look at $\ x^2+6x+10 \$..While you're at it, consider any problems associated with $\ 2x^2 + 2x + 3 \$ as the base for a logarithm.