Identities for solving log questions

[Saitama]

In an equality like:
a / b > 0
there are 2 ways to satisfy the inequality.
Either a and b both have to be positive, or a and b both have to be negative.

Since the denominator (b) is a square root, we already know that b >= 0.

All clear so far?

Yes, clear

 

[Saitama]

Please explain me further

 

[I like Serena]

Please explain me further

Rather than repeat what the solution key already states, can you tell me what you don't understand?

Perhaps what you don't realize yet, is that when you have a function like a square root or a log function, it is not defined for any value.

A square root is *only* defined for numbers greater than or equal to zero.
So this limits possible solutions.

Similarly a quotient like a / b is *only* defined if b is not equal to zero.

And finally a log function can only be taken from a number that is greater than zero.

 

[Saitama]

I am not able to understand these statements:-

The trinomial x²-4x+11 = (x-2)³ + 7 $\ge$ 7 for any x $\in$ R, and in that case log₅ (x²-4x+11)² > 0.

I want to know how this relation came out:-
x²-4x+11 = (x-2)³ + 7 $\ge$ 7

 

[I like Serena]

I am not able to understand these statements:-

The trinomial x²-4x+11 = (x-2)³ + 7 $\ge$ 7 for any x $\in$ R, and in that case log₅ (x²-4x+11)² > 0.

I want to know how this relation came out:-
x²-4x+11 = (x-2)³ + 7 $\ge$ 7

This is called "completing the square".
It's a method to solve a quadratic equation.
A detailed explanation can be found on wikipedia:
http://en.wikipedia.org/wiki/Completing_the_square

In this case it's used to show that the expression is always greater than zero, which can also be shown in other ways.

Can you see that the following expressions are equal?
x²-4x+11 = (x-2)² + 7

Going on from there, a squared number is always greater or equal to zero, since the square of a negative number is positive.
Adding 7 makes only extra sure that it is greater than zero.

 

[Saitama]

Git it :) Thanks!

But what does this statement mean?

Thus, for left-hand side of the inequality to have sense, it is necessary that both inequalities be satisfied, i.e. that x $\in$ (-2, 2-$\sqrt{15}$).

Which inequalities is it talking about?

 

[I like Serena]

Git it :) Thanks!

But what does this statement mean?

Thus, for left-hand side of the inequality to have sense, it is necessary that both inequalities be satisfied, i.e. that x $\in$ (-2, 2-$\sqrt{15}$).

Which inequalities is it talking about?

The argument to the square root in the denominator must be greater than zero.
And the argument to the log function must be greater than zero.

 

[Saitama]

So if i take different cases then i will be able to solve the question. Am i right?

 

[SammyS]

I am not able to understand these statements:-

The trinomial x²-4x+11 = (x-2)³ + 7 $\ge$ 7 for any x $\in$ R, and in that case log₅ (x²-4x+11)² > 0.

I want to know how this relation came out:-
x²-4x+11 = (x-2)³ + 7 $\ge$ 7

There's a typo here.

It should be: x²-4x+11 = (x-2)² + 7 .

(x-2) is squared, not cubed.

 

[SammyS]

Ok i am writing everything what is given in the answer key:-

"Since the denominator of the fraction is always positive, the numerator must be $\ge$ 0 for the inequality to be satisfied. The trinomial 2-5x-x² > 0 for all x $\in$ (-2,3), and the denominator is meaningful only for these values. The trinomial x²-4x+11 = (x-2)² + 7 $\ge$ 7 for any x $\in$ R, and in that case log₅ (x²-4x+11)² > 0. The trinomial x²-4x-11 assumes nonnegative values for x $\in$ (-$\infty$, 2-$\sqrt{15}$) $\cup$ (2+$\sqrt{15}$, +$\infty$), and only for these values log₁₁ (x²-4x-11)³ is meaningful. Thus, for left-hand side of the inequality to have sense, it is necessary that both inequalities be satisfied, i.e. that x $\in$ (-2, 2-$\sqrt{15}$). On that interval f(x) = log₁₁ (x²-4x-11) < 0. Indeed, f(-2) = log₁₁ 1³ = 0, f(2-$\sqrt{15}$) = log₁₁ 0 = -$\infty$. The trinomial x²-4x-11 attains its minimum value for x=2. When varies to the left of x=2, the values of the trinomial increase continuously, for x=2-$\sqrt{15}$ its value is equal to zero, for x = +2 it is equal to 1; thus f(x) increase monotonically when x varies from 2-$\sqrt{15}$ to -2, remaining negative all the time. Consequently, on the left hand side of the inequality the numerator of the fraction is positive for x $\in$ (-2, 2-$\sqrt{15}$) and the inequality is valid for these values."

Would someone please explain this to me?

The above is rather longer than it needs to be. Parts of it are downright confusing.

For the inequality you're working with, the important issue is the domain of the expression on left hand side:$$\frac{\log_5((x^2-4x+11)^2)-\log_{11}((x^2-4x-11)^3)}{\sqrt{2-5x-x^2}}$$

The domain for log₅ (x²-4x+11)² is all real numbers.

The domain for log₁₁ (x²-4x-11)³ is $(-\infty,\,2-\sqrt{15})\cup(2+\sqrt{15},\,+\infty)\,.$

The domain for $1/\sqrt{2-5x-x^2}$ is the interval: (-2, 3) .

Put these together to get the domain: $(-2,\,2-\sqrt{15}))$ ≈ (-2, -1.87...) .

Furthermore, the polynomial (x²-4x-11)³, is monotone decreasing on (-∞, 2), so it's monotone decreasing on $(-2,\,2-\sqrt{15}))$.

x²-4x-11 = 1 at x = -2.
x²-4x-11 = 0 at x = $2-\sqrt{15}$.

So, 1 > x²-4x-11 > 0 on $(-2,\,2-\sqrt{15}))$.​

The above tells you that log₁₁ (x²-4x-11)³ is negative for this entire interval. Note: log₁₁(a) is negative if 1 < a < 0.

Altogether, this tells you that the left hand side of the inequality is positive over its entire domain.

Added in Edit.

There are a couple of typos above. That should have been: The domain for $1/\sqrt{2-5x-3x^2}$ is the interval: (-2, 3) .

Of course, the correct inequality is: $$\frac{\log_5((x^2-4x+11)^2)-\log_{11}((x^2-4x-11)^3)}{\sqrt{2-5x-3x^2}}\ge0$$

The rest is OK. (I think.)

 

[Saitama]

The domain for $1/\sqrt{2-5x-x^2}$ is the interval: (-2, 3) .

It isn't $\sqrt{2-5x-x^2}$, it is $\sqrt{2-5x-3x^2}$...

 

[SammyS]

It isn't $\sqrt{2-5x-x^2}$, it is $\sqrt{2-5x-3x^2}$...

I corrected my post.

 

[Saitama]

I corrected my post.

But when you find the domain interval of 2-5x-3x², it isn't (-2,3)..

 

[SammyS]

But when you find the domain interval of 2-5x-3x², it isn't (-2,3)..

You're right.

The domain of $\sqrt{2-5x-3x^2}$ is [-2, 1/3].

So, the domain of $1/\sqrt{2-5x-3x^2}$ is (-2, 1/3).

However, that does give the same the domain, $(-2,\,2-\sqrt{15}))$, for the overall expression.

I apologize. I should have been more careful in composing my earlier posts.
In my defense, those errors come from what appear to be typos in the post that displayed the answer key. That's not a very good excuse on my part. I should have been more careful. (Too much cutting & pasting without checking!)​

 

[Saitama]

You're right.

The domain of $\sqrt{2-5x-3x^2}$ is [-2, 1/3].

So, the domain of $1/\sqrt{2-5x-3x^2}$ is (-2, 1/3).

However, that does give the same the domain, $(-2,\,2-\sqrt{15}))$, for the overall expression.

I apologize. I should have been more careful in composing my earlier posts.
In my defense, those errors come from what appear to be typos in the post that displayed the answer key. That's not a very good excuse on my part. I should have been more careful. (Too much cutting & pasting without checking!)​

Thanks Sammy for solving my question. :)
I will work on it and if i find any doubts, i would post it here :)...

And no need to apologize because the hint itself is too confusing. When i read it first, i too didn't noticed the typo...

 

[SammyS]

These graphs (made by WolframAlpha) may help.

 

[I like Serena]

Hi Pranav-Arora!

No more questions?
I'm surprised!
I had expected another problem by now!

 

[Saitama]

Hi Pranav-Arora!

No more questions?
I'm surprised!
I had expected another problem by now!

There are more questions but i was trying to again solve them before posting....
You will soon be getting more questions by me...

 

[Saitama]

So here's my next question:-

$$log_\frac{1}{\sqrt{5}}(6^{x+1} - 36^x) \ge -2$$

 

[I like Serena]

Hi Pranav-Arora!

So here's my next question:-

$$log_\frac{1}{\sqrt{5}}(6^{x+1} - 36^x) \ge -2$$

So did you try anything?

 

[Saitama]

So did you try anything?

Yep, i substituted y at the place of 6^x and got this:-

6y-y² $\le$ 5

I solved it and got [0, log₆ 5]
but the answer in the answer key is (-$\infty$,0]$\cup$[log₆ 5,1)...

 

[I like Serena]

Yep, i substituted y at the place of 6^x and got this:-

6y-y² $\le$ 5

I solved it and got [0, log₆ 5]
but the answer in the answer key is (-$\infty$,0]$\cup$[log₆ 5,1)...

Seems as if you understand quite a lot already!
Just a few details left...

What's the difference between your answer and the answer key?
That is, what should change in your equation to make it match the answer key more or less?
EDIT: Scratch that, how does your equation mismatch the answer key?

Also, could you show a couple of intermediary steps?

 

[Mentallic]

Yep, i substituted y at the place of 6^x and got this:-

6y-y² $\le$ 5

I solved it and got [0, log₆ 5]

This problem isn't quite as simple as that.

If you have x>y, then a^x>a^y only if a>1. If 0<a<1 then we need to switch the inequality. It seems like you did this already but then you ended up with an answer that would assume you didn't? Anyway, $$a=1/\sqrt{5}<1$$ in this case so you'll need to switch it.

The second problem is you haven't considered when the original question is valid. Remember that you can only take the log of a positive value.

 

[Saitama]

The second problem is you haven't considered when the original question is valid. Remember that you can only take the log of a positive value.

I have switched the inequality sign and you are right, i didn't took this case:-

$(6^{x+1} - 36^x) > 0$

Also, could you show a couple of intermediary steps?

I did a step wrong...
So i will correct it here:- (If there any mistakes, please tell me)

6y-y2 ≤ 5
$\Rightarrow$y²-6y+5 > 0
$\Rightarrow$(y-1)(y-5) > 0
$\Rightarrow$y $\in$ (-$\infty$, 1]$\cup$[5, $\infty$)
$\Rightarrow$6^x $\in$ (-$\infty$, 1]$\cup$[5, $\infty$)

From this how i will find the value of x?

 

[I like Serena]

From this how i will find the value of x?

You set y = 6^x.

So x = log₆ y

but of course this is only true for y > 0, limiting the solutions for x!


Oh, and you included the boundaries 1 and 5 in your solution set for y, but these should be excluded!
EDIT: Scratch that. I keep making mistakes here, I'm sorry to say. You changed ≤ into >, but that is not proper, because the equality is still possible.

 

[Saitama]

You set y = 6^x.

So x = log₆ y

but of course this is only true for y > 0, limiting the solutions for x!


Oh, and you included the boundaries 1 and 5 in your solution set for y, but these should be excluded!

Do i need to take this case?
6y-y² > 0

If so, after finding the domain interval, what should i do?

 

[I like Serena]

Do i need to take this case?
6y-y² > 0

If so, after finding the domain interval, what should i do?

Yes, you need to take this case and find the solution set.
You also have another solution set for the complete equation.

Those need to combined, that is, you need to take the intersection of both solution sets.
That is, because a solution can only be a solution, if when you fill it in in the original equation, all domain-requirements of the functions you use must be satisfied and the entire equation must be satisfied.

Btw, I edited my previous post, since I made a mistake there.
I'll try to be more careful from now on!

 

[Saitama]

Yes, you need to take this case and find the solution set.
You also have another solution set for the complete equation.

Those need to combined, that is, you need to take the intersection of both solution sets.
That is, because a solution can only be a solution, if when you fill it in in the original equation, all domain-requirements of the functions you use must be satisfied and the entire equation must be satisfied.

Btw, I edited my previous post, since I made a mistake there.
I'll try to be more careful from now on!

I solved this case:-

6y-y² $\ge$ 5

but how would i solve 6^x = -$\infty$?

 

[I like Serena]

Hi again Pranav-Arora!

I solved this case:-

6y-y² $\ge$ 5

but how would i solve 6^x = -$\infty$?

This has no solution, since 6^x is always positive! Actually, you solved:

6y-y² ≤ 5
⇒ y ∈ (-∞, 1]∪[5, ∞)

So your solution set for y is (-∞, 1]∪[5, ∞).

To find the solution set for x, you need to calculate x = log₆ y, but only for those y for which log₆ is defined.
That is for the set (0, 1]∪[5, ∞).

That would be: (-∞, log₆ 1]∪[log₆ 5, ∞)

 

[Saitama]

Thanks!It worked!

My next question:-

$$\log_3\frac{|x^2-4x|+3}{x^2+|x-5|} \ge 0$$