Identities for solving log questions

[Saitama]

If you multiply left and right by the denominator, the inequality will swap around if the denominator is negative.

(Note that this also plays a role in your first question! )

I did as you said.
I first took x>1 and tried to solve it.
I got:-

\log_2 (x+10) * (x-1) > 0

So what's the next step...

 

[I like Serena]

I did as you said.
I first took x>1 and tried to solve it.
I got:-

\log_2 (x+10) * (x-1) > 0

So what's the next step...

Hmm, what you should have is:
\log_2 (x+10) > 0

And from there you get:
x+10 > 2⁰

So:
x > -9

Since we restricted ourselves in this case to x > 1, that just leaves:
x > 1

Now we need to combine this result to what happens if x < 1 ...

 

[Saitama]

lol...I didnt canceled (x-1) while multiplying both the sides with (x-1)...

I solved it but i am getting the answer to be (1,infinity) but when i checked the answer key it is (-1,0) U (1,infinity)...:(

I solved it four times but then too got the same answer...
Please help...

 

[I like Serena]

lol...I didnt canceled (x-1) while multiplying both the sides with (x-1)...

I solved it but i am getting the answer to be (1,infinity) but when i checked the answer key it is (-1,0) U (1,infinity)...:(

I solved it four times but then too got the same answer...
Please help...

What's the solution if x < 1?

Note that multiplying by (x-1) in this case inverts the inequality.

 

[Saitama]

I am very sorry for my foolishness...:(
It wasn't "10", it was "1" but the method you gave worked successfully for my question...
Thank you very much!

I wanted to know that what is x when 2^x > 0.
I solved it using the reverse of anti-log method and i got x > log₂ 0.
So what should i do next?

 

[Mentallic]

I am very sorry for my foolishness...:(
It wasn't "10", it was "1" but the method you gave worked successfully for my question...
Thank you very much!

I wanted to know that what is x when 2^x > 0.
I solved it using the reverse of anti-log method and i got x > log₂ 0.
So what should i do next?

Think about the shape and details of any exponential graph.

 

[Saitama]

Think about the shape and details of any exponential graph.

Sorry Mentallic, i don't know about exponential graphs..:(

 

[HallsofIvy]

You still don't say what the question is, but I will guess that it is to find x so that the inequality is satisfied.

The product of two numbers is negative if and only if the two numbers are of opposite sign. So your problem reduces to

Case I:
log_4\left(\frac{x+ 1}{x+2}\right)&lt; 0
and
log_4\left(x+ 3)&gt; 0

or
Case II
log_4\left(\frac{x+ 1}{x+2}\right)&gt; 0
and
log_4\left(x+ 3)&lt; 0

 

[Mentallic]

Sorry Mentallic, i don't know about exponential graphs..:(

You're solving logarithmic equations but you don't know what an exponential graph looks like? The latter should definitely come first in your studies...

Anyway, think about what the value of 2^x is for large positive x, small positive x, small negative x and large negative x.

 

[I like Serena]

I wanted to know that what is x when 2^x > 0.
I solved it using the reverse of anti-log method and i got x > log₂ 0.
So what should i do next?

Uhm, I'm not sure what you're asking here, so here's from the top of my head.

2^x > 0
This is true for any x.

x > log₂ 0
Yes this is equivalent (sort of), and since log₂ 0 = -∞, this means this holds for any x.

 

[Saitama]

Uhm, I'm not sure what you're asking here, so here's from the top of my head.

2^x > 0
This is true for any x.

x > log₂ 0
Yes this is equivalent (sort of), and since log₂ 0 = -∞, this means this holds for any x.

Thanks! When i used log₂ 0= -infinity , my answer matched the answer in the answer key.
But when i was solving a problem i got 2^x>-infinity. I applied the same process and got x > log₂ -infinity. But log of negative number is not possible, so what should i do next?

 

[I like Serena]

Thanks! When i used log₂ 0= -infinity , my answer matched the answer in the answer key.
But when i was solving a problem i got 2^x>-infinity. I applied the same process and got x > log₂ -infinity. But log of negative number is not possible, so what should i do next?

There's no need to take a log in this case.
What you need to know is that 2^x > 0 is always true for any x.

 

[Saitama]

There's no need to take a log in this case.
What you need to know is that 2^x > 0 is always true for any x.

Sorry! But i didn't get you...

 

[I like Serena]

Sorry! But i didn't get you...

Do you get me now?

 

[Saitama]

Do you get me now?

No i didnt get you! I am asking how to solve for x when 2^x>-infinity...

 

[I like Serena]

No i didnt get you! I am asking how to solve for x when 2^x>-infinity...

Hmm, I'm trying to explain there's nothing to solve.
The solution set is the set of all real numbers.

How much do you know about exponentiation?
I take it you know that 2¹ = 2 and 2² = 4.

Do you also know that 2^-1 = 1/2 and 2^-2 = 1/4?

Here's a graph of an exponentiation function (actually e^x):

Note that the graph is entirely above the x axis.
So 2^x > 0 for any x.
And by implication 2^x > - infinity for any x.

 

[Saitama]

Hmm, I'm trying to explain there's nothing to solve.
The solution set is the set of all real numbers.

How much do you know about exponentiation?
I take it you know that 2¹ = 2 and 2² = 4.

Do you also know that 2^-1 = 1/2 and 2^-2 = 1/4?

Here's a graph of an exponentiation function (actually e^x):

Note that the graph is entirely above the x axis.
So 2^x > 0 for any x.
And by implication 2^x > - infinity for any x.

Thanks! I got it! So that means x is an element of all real numbers...

 

[Saitama]

I am again getting a problem in this question:-

I solved it as you said I like Serena but i got stuck:-
x^2-4x+11 \ge (x^2-4x-11)^\frac 3 {2 \log_5 11}

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

 

[I like Serena]

I am again getting a problem in this question:-

I solved it as you said I like Serena but i got stuck:-
x^2-4x+11 \ge (x^2-4x-11)^\frac 3 {2 \log_5 11}

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

Mathematically speaking there's not much else you can do.
You can't solve this algebraically, but only numerically.

For values of x close to zero, the inequality holds.
For large positive x (starting around 20) or negative x (starting around -20), the inequality will not hold.
You can only find an approximation of these values.

 

[Saitama]

Mathematically speaking there's not much else you can do.
You can't solve this algebraically, but only numerically.

For values of x close to zero, the inequality holds.
For large positive x (starting around 20) or negative x (starting around -20), the inequality will not hold.
You can only find an approximation of these values.

When i checked out the answer key, i found the answer to be (-2,2- \sqrt{15} )...
Would you please tell me how can i get this answer?

I don't understand why square root of 15 doesn't come in line with 2

 

[I like Serena]

When i checked out the answer key, i found the answer to be (-2,2- \sqrt{15} )...
Would you please tell me how can i get this answer?

I don't understand why square root of 15 doesn't come in line with 2

Here you can view the results of the equation:
http://www.wolframalpha.com/input/?i=%28log_5%28%28x^2-4x%2B11%29^2%29+-+log_11%28+%28x^2-4x-11%29^3%29%29+%2F+sqrt%282-5x-3x^2%29%3D0

So the approximate boundaries for x are -24 and +20.
This does not match the solution you just gave.

I already thought this problem was more complex than seemed likely seeing the difficulty level of the other problems you gave.
Is there perhaps a typo in your problem?


To get sqrt 15 in line, use either [ itex ] instead of [ tex ], or use the sqrt symbol √ (without tex).

 

[Saitama]

Here you can view the results of the equation:
http://www.wolframalpha.com/input/?i=%28log_5%28%28x^2-4x%2B11%29^2%29+-+log_11%28+%28x^2-4x-11%29^3%29%29+%2F+sqrt%282-5x-3x^2%29%3D0

So the approximate boundaries for x are -24 and +20.
This does not match the solution you just gave.

I already thought this problem was more complex than seemed likely seeing the difficulty level of the other problems you gave.
Is there perhaps a typo in your problem?To get sqrt 15 in line, use either [ itex ] instead of [ tex ], or use the sqrt symbol √ (without tex).

Yes there is a typo. Its log₁₁ (x²-4x-11)³.
But i already mentioned it before.

 

[I like Serena]

Yes there is a typo. Its log₁₁ (x²-4x-11)³.
But i already mentioned it before.

Yes, I already accounted for that.
And I just edited my previous post so you can simply click on the WolframAlpha link.

 

[Saitama]

Yes, I already accounted for that.
And I just edited my previous post so you can simply click on the WolframAlpha link.

So what's the solution?

 

[I like Serena]

So what's the solution?

Looking only at the difference in log functions, this would be the set [ -20.9021, +24.9021 ].

However, we neglected to look at the square root denominator, because to have a proper solution, the expression within the square root must be greater than zero.
So we still need to solve:

2-5x-3x² > 0

Do you know the solution to that one?

 

[Saitama]

I think the answer shouldn't be -2,2-\sqrt{15} because when i checked the answer key, a hint is given in the form of a big paragraph. In that, it is written "The trinomial 2-5x-x² >0 for all x \in (-2,3) and the denominator is only meaningful only for these values." But in the question it is given 2-5x-3x². So now we are sure that the answer isn't -2, 2-\sqrt{15}.

And i can solve 2-5x-3x²>0 but would you please tell me how you got -20.9021, +24.9021.

Would you please also tell me what's the difference between these two statements (log_a b)² and log_a² b.

@I like Serena:-Also, just wanted to ask you that are you having this book "Problems in Mathematics with Hints and Solutions" by "V Govorov, P Dybov, N Miroshin & S Smirnova" and edited by "Prof. A.I. Prilepko" because i am asking these questions from this book?

 

[I like Serena]

And i can solve 2-5x-3x²>0 but would you please tell me how you got -20.9021, +24.9021.

Since we can't solve it algebraically, we need an approximation.
You can see it here:
http://www.wolframalpha.com/input/?i=%28log_5%28%28x^2-4x%2B11%29^2%29+-+log_11%28+%28x^2-4x-11%29^3%29%29%3D0

I have let Wolfram Alpha do the approximation for me.

Btw, there is yet another complication, which I missed before.
Since we have log_{11}( (x^2-4x-11)^3) we have the additional constraint that x^2-4x-11 &gt; 0 , because you can't take a log function from a negative number.
You'll have to solve that one as well.

Would you please also tell me what's the difference between these two statements (log_a b)² and log_a² b.

No difference, the second form is a short hand notation of the first.

@I like Serena:-Also, just wanted to ask you that are you having this book "Problems in Mathematics with Hints and Solutions" by "V Govorov, P Dybov, N Miroshin & S Smirnova" and edited by "Prof. A.I. Prilepko" because i am asking these questions from this book?

Sorry, but no, I don't have this book.
I don't have any books that explain your material.

 

[SammyS]

You can only take the log of a positive number, so you must have: x²-4x-11 > 0 . For what values of x is that true?

Similarly, you can only take the square root of a non-negative number. Also, the square root is in the denominator, so you must also have: 2-5x-3x²>0 . For what values of x is this true?

It looks to me like the answer key is correct !

Look at this from WolframAlpha. (Give it extra computing time.)

 

[Saitama]

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)³ + 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?

 

[I like Serena]

What is it that you do not understand?
I would have to guess at what to explain.
Can you perhaps show what you do understand?

Let's break the solution down.
I'll start with the first sentence.

"Since the denominator of the fraction is always positive, the numerator must be \ge 0 for the inequality to be satisfied.

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?