Identities for solving log questions

[Saitama]

I was wondering if there could be more identities than i have read. I was doing questions on log and found many questions in this form:-

(log_a b)²

(log_a b)(log_a c)

This is not my homework but i require these identities to solve the questions.
Please someone tell me how to solve these type of questions.

Thanks.

 

[QuarkCharmer]

http://en.wikipedia.org/wiki/List_of_logarithmic_identities

 

[praharmitra]

I don't believe there are any identities for these quantities.

 

[I like Serena]

I don't believe there are any identities for these quantities.

True!

 

[Saitama]

http://en.wikipedia.org/wiki/List_of_logarithmic_identities

I already checked them out before.
But I am having questions of this type, how would i solve them?

 

[I like Serena]

I already checked them out before.
But I am having questions of this type, how would i solve them?

Do you have a specific question?

 

[Mentallic]

I already checked them out before.
But I am having questions of this type, how would i solve them?

Well since the question isn't "use a log identity to simplify (log_ab)² then they aren't exactly of this type now are they.

 

[Saitama]

Ok, so no identities of this type exist...
But is it possible to simplify them?

 

[I like Serena]

Ok, so no identities of this type exist...
But is it possible to simplify them?

Only if you know the numbers, then you can use a calculator.

 

[Saitama]

The questions which i have to solve include variable so i can't use a calculator...:(

 

[Mentallic]

The questions which i have to solve include variable so i can't use a calculator...:(

If you give us the question we'll be better equipped to help you.

 

[Saitama]

Here's a question:-

Now how would i solve (log_e x)²

 

[I like Serena]

Here's a question:-

Now how would i solve (log_e x)²

Substitute y=log_e(x) and solve for y.

When you're done (if you find a solution), back substitute x = e^y.

 

[Saitama]

Substitute y=log_e(x) and solve for y.

When you're done (if you find a solution), back substitute x = e^y.

Wow! Why i didnt just used my head?
It worked...

One more question, this time in the form (log_a b)(log_a c):-

 

[I like Serena]

What's the question?

 

[Saitama]

What's the question?

Edited!

 

[I like Serena]

Edited!

This is of the form:

a x b < 0​

The solution is:

a < 0 and b > 0​

or

a > 0 and b < 0​

Solve from there!

 

[Saitama]

May i know why is

a < 0 and b > 0
or
a > 0 and b < 0

 

[I like Serena]

May i know why is

a < 0 and b > 0
or
a > 0 and b < 0

If you multiply two positive numbers, the result is positive.
If you multiply two negative numbers, the result is positive.
If one of the two numbers is zero, the result is zero.

So one number must be positive and the other must be negative.

 

[Saitama]

If you multiply two positive numbers, the result is positive.
If you multiply two negative numbers, the result is positive.
If one of the two numbers is zero, the result is zero.

So one number must be positive and the other must be negative.

Thanks, it worked...

I am having loads of doubts and questions in log. Can i post them here one by one?

 

[I like Serena]

Thanks, it worked...

I am having loads of doubts and questions in log. Can i post them here one by one?

Sure. :)
That way it'll bump my attention.

Note that if I'm not available, you'll get a quicker response if you create a new thread.
Up to you though.

 

[Saitama]

Thanks..:)
So my next question:-

I solved it but i got stuck at:-

 

[I like Serena]

Thanks..:)
So my next question:-

I solved it but i got stuck at:-

How about multiplying left and right with the square root?
Note that a square root is always greater than or equal to zero.

Furthermore the base powers of the logs are different. That would make the equation a bit harder.
Are they different?
Because you applied the log-identities as if they are the same.

EDIT: Apparently you extracted the power of 2 inconsistently, because it should cancel.

 

[Saitama]

Sorry its log₁₁ (x²-4x-11)³

(I am solving the numerator part)
I changed the base log11 to log5.
It came out to be like this:-

log₅ (x2-4x-11)
----------------
log₅ 11

log₅ 11 is around 1.5.
Then i got the numerator:-

- 3log₅ (x²-4x-11) + 2log₅ (x²-4x+11)
-----------------
1.5
then reduced three to 2 since 1.5 *2 = 3
and then took 2 outside...

and i got the result where i am now stuck.

 

[Saitama]

@I like Serena:- Didn't you see my reply?

 

[I like Serena]

@I like Serena:- Didn't you see my reply?

No, I didn't see your reply.

This time round I did not get a notification from either of your last 2 replies.
I don't know why, although I saw other threads mentioning that apparently the PF notification mechanism does not always work.

It's by coincidence that I happened to see in the forum overview that you posted new replies.

Sorry its log₁₁ (x²-4x-11)³

(I am solving the numerator part)
I changed the base log11 to log5.
It came out to be like this:-

log₅ (x2-4x-11)
----------------
log₅ 11

log₅ 11 is around 1.5.
Then i got the numerator:-

- 3log₅ (x²-4x-11) + 2log₅ (x²-4x+11)
-----------------
1.5
then reduced three to 2 since 1.5 *2 = 3
and then took 2 outside...

and i got the result where i am now stuck.

Let's not make approximations until we have to.
You'll get the wrong answer because of it.

If you work it out you should find:

2 \log_5 \frac {x^2-4x+11} {(x^2-4x-11)^\frac 3 {2 \log_5 11}} \ge 0

This means:

x^2-4x+11 \ge (x^2-4x-11)^\frac 3 {2 \log_5 11}

You should see that the power on the right hand side is slightly bigger than the one on the left hand side, meaning the inequality won't hold for large x.

Picking a large x and substituting it in your expression should show that the original inequality does not hold either.

 

[Saitama]

No, I didn't see your reply.

This time round I did not get a notification from either of your last 2 replies.
I don't know why, although I saw other threads mentioning that apparently the PF notification mechanism does not always work.

It's by coincidence that I happened to see in the forum overview that you posted new replies.




Let's not make approximations until we have to.
You'll get the wrong answer because of it.

If you work it out you should find:

2 \log_5 \frac {x^2-4x+11} {(x^2-4x-11)^\frac 3 {2 \log_5 11}} &gt; 0

This means:

x^2-4x+11 \ge (x^2-4x-11)^\frac 3 {2 \log_5 11}

You should see that the power on the right hand side is slightly bigger than the one on the left hand side, meaning the inequality won't hold for large x.

Picking a large x and substituting it in your expression should show that the original inequality does not hold either.

Thanks it worked! You are really great!

My next question:-
\frac{\log_2 (x+10}{x-1} &gt; 0

(Sorry for the late reply..:(
I had been away for a few days)

 

[I like Serena]

Thanks it worked! You are really great!

My next question:-
\frac{\log_2 (x+10)}{x-1} &gt; 0

(Sorry for the late reply..:(
I had been away for a few days)

Uhh? What is your problem?

You'd have to consider the 2 cases where x > 1 and where x < 1 separately, but from there it should be straight forward?

 

[Saitama]

Uhh? What is your problem?

You'd have to consider the 2 cases where x > 1 and where x < 1 separately, but from there it should be straight forward?

There's a closed bracket after 10.

And why i need to take these two cases?

 

[I like Serena]

There's a closed bracket after 10.

And why i need to take these two cases?

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! )

 

[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) &gt; 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) &gt; 0

So what's the next step...

Hmm, what you should have is:
\log_2 (x+10) &gt; 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