Identities for solving log questions

[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:-

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

 

[I like Serena]

Hi Pranav-Arora!

Thanks!It worked!

My next question:-

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

Hmm, that's not a question - it's a problem.
What is your question?

I suspect that the log should bother you no more by now, nor the matter of the domain of the log function.

The only new thing is the absolute value function...
And I'm not sure whether you fully understand when an inequality inverts and how exactly...

Perhaps show a few steps...?

 

[Saitama]

Hi Pranav-Arora!



Hmm, that's not a question - it's a problem.
What is your question?

I suspect that the log should bother you no more by now, nor the matter of the domain of the log function.

The only new thing is the absolute value function...
And I'm not sure whether you fully understand when an inequality inverts and how exactly...

Perhaps show a few steps...?

When i tried to solve it, i wasn't able to go further than this:-

$$\frac{|x^2-4x|+3}{x^2+|x-5|} \ge 1$$

I am not able to figure out what should i do next?

 

[I like Serena]

When i tried to solve it, i wasn't able to go further than this:-

$$\frac{|x^2-4x|+3}{x^2+|x-5|} \ge 1$$

I am not able to figure out what should i do next?

What would happen if you multiplied left and right by the denominator?

Can you separate this in a number of cases for x?
That is, for instance, what would the signs of the absolute values be if x > 5?
For which values of x would one or the other absolute value invert its sign?

 

[I like Serena]

When i tried to solve it, i wasn't able to go further than this:-

$$\frac{|x^2-4x|+3}{x^2+|x-5|} \ge 1$$

I am not able to figure out what should i do next?

What would happen if you multiplied left and right by the denominator?

Can you separate this in a number of cases for x?
That is, for instance, what would the signs of the absolute values be if x > 5?
For which values of x would one or the other absolute value invert its sign?

 

[Saitama]

What would happen if you multiplied left and right by the denominator?

Can you separate this in a number of cases for x?
That is, for instance, what would the signs of the absolute values be if x > 5?
For which values of x would one or the other absolute value invert its sign?

Absolute values are positive only...so why are you asking me "what would the signs of the absolute values be if x > 5?"

 

[I like Serena]

Absolute values are positive only...so why are you asking me "what would the signs of the absolute values be if x > 5?"

What I meant is that you can write an absolute value without the vertical bars if you put the appropriate sign before it in combination with a domain restriction.

For instance:
|x-5| = (x-5) if x > 5
|x-5| = -(x-5) if x < 5

 

[ehild]

Look at the denominator. Can it be negative?

ehild