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#201 2013-08-04 05:05:28

bobbym

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Re: Simplify the following:

Sorry but I can not make any sense out of the second one. There are just too many ways to group that. Please group it.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#202 2013-08-04 07:26:34

EbenezerSon
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Re: Simplify the following:

But that was how it exists in the book and I tried to solve.
Think the following is how it came by its answder.

Logy^2-log2x=log2y-log2x

(y^2/2x)= 2x(2y-2x).

y^2= 4xy-4x^2

y^2= 4xy-4x^2

y^-4xy+4x^2=0 I think could be factorised,

My question is,   as there became a negative sign between log2y-log2x I think should be log(2y/2x). As the law says logx-logy=(x/y).
But why only the law affected the left hand equation  only I mean (y^2/2x) and not the right hand as well?

Last edited by EbenezerSon (2013-08-04 08:00:14)

#203 2013-08-04 08:25:07

EbenezerSon
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Re: Simplify the following:

I know the book is right, but I can not figure out as to why it applied the law of division on the left equation only. I could see the right equation also has a negative sign between it and I think should be
lo2y/log2x so that it would also have a division sign as the left equation has. Then from there one can work it down.
Bobbym, why do you think the division sign was denied on the right hand equation only?
Or is only left hand equations  that must assume it? And not right hand equations?

#204 2013-08-04 09:25:18

anonimnystefy
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Re: Simplify the following:

Notice that your right hand side isn't the same as in the original problem from post #194.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#205 2013-08-04 14:58:04

bobbym

Online

Re: Simplify the following:

That book has lots of mistakes in it:

Last but not the least
log81/log3^1.
the  book has -4 at its back as the answer.

I am getting 4 not -4.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#206 2013-08-04 21:51:35

zetafunc.
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Re: Simplify the following:

Which textbook are these questions from?

#207 2013-08-04 21:52:30

bobbym

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Re: Simplify the following:

I have not asked him that yet.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#208 2013-08-05 00:57:42

EbenezerSon
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Re: Simplify the following:

Anonynmstify, if I am getting you right I think youre right, but then when you work it down you would arrive on the division sign, watch:

2logy-log2x=log2(y-x).

logy^2 - log2x = log2y-2x. Now, there come a negative sign at the right equation.
I am thinking that, since there is a negative sign in both the right and the left equation, both sides must have that division sign respectively. But to my suprise only the left equation had the division sign. Still I am not clear.
Bobym let me know why +4.

#209 2013-08-05 01:09:27

bobbym

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Re: Simplify the following:

Last but not the least
log81/log3^1.
the  book has -4 at its back as the answer.

Now just cancel the log(3) on the top and bottom.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#210 2013-08-05 01:16:20

EbenezerSon
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Re: Simplify the following:

Bobym, please could you produce the methods here, if the log3^1 were to be log1/3. What would be the answer?

Thanks.

#211 2013-08-05 01:19:52

bobbym

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Re: Simplify the following:

The method is in post #209, I showed the three steps.

If the denominator was 1 / 3 instead of 3 then the answer would be -4.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#212 2013-08-05 01:34:22

EbenezerSon
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Re: Simplify the following:

Thanks, Bobbym!
God bless.

1/3logp=1. Find the value of p.

Bobbym, how would one do this? I have tried and my answer was not the same as what the book has.

#213 2013-08-05 01:36:41

anonimnystefy
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Re: Simplify the following:

p=1000.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#214 2013-08-05 01:45:16

bobbym

Online

Re: Simplify the following:

1/3logp=1

Times both sides by 3

log(p) = 3

Take 10^ of both sides.

p = 10^3 = 1000

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#215 2013-08-05 01:59:54

EbenezerSon
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Re: Simplify the following:

See, I am taking aback as to why the answer is 1000.
Because I have learnt that any number raised to power either 1/3, 1/2, 1/4. is demanding the third the square and the fourth root of that number. I see it that if the answer is 1000  in the  above problem  then the question should have been:

3logp=1. so that it would be,

log p^3=log10. taking antilog.

p^3 = 10
p =10^3.

p=1000

Please, anyone correct me if I am wrong.

Thanks.

#216 2013-08-05 02:13:25

EbenezerSon
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Re: Simplify the following:

I mean

1/3logp=1
logp^1/3=10
p =10^1/3
third root is required.
p=2.15.

I am not saying anyone is wrong but this suprise me, c'os I have applied what I am talking about for long.

#217 2013-08-05 02:15:27

bobbym

Online

Re: Simplify the following:

Line 3 is incorrect.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#218 2013-08-05 02:36:53

EbenezerSon
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Re: Simplify the following:

Okay, then is the following correct?

For instance.

y^1/3=a. would be a^
3
y^3 = a. square root of a would then be taken.
Is the above right?

#219 2013-08-05 02:38:31

bobbym

Online

Re: Simplify the following:

First one is okay.

y^3 = a. square root of a would then be taken.

You mean cube root.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#220 2013-08-05 02:45:29

EbenezerSon
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Re: Simplify the following:

But are they all right in that manner?

#221 2013-08-05 02:54:52

bobbym

Online

Re: Simplify the following:

Yes, they are correct.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#222 2013-08-05 04:50:59

anonimnystefy
Real Member

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Re: Simplify the following:

Have you sorted out the x and y problem?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#223 2013-08-05 04:52:39

bobbym

Online

Re: Simplify the following:

What x and y problem?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#224 2013-08-05 05:29:20

anonimnystefy
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Re: Simplify the following:

EbenezerSon wrote:

If 2logy-log2x=2log(y-x), express y in terms of x. I had (y-x)(y-2x^2). The book has y^-4xy+4^2x=0.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#225 2013-08-05 07:56:52

EbenezerSon
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Re: Simplify the following:

I have solved it but this way:
2logy-log2x =  log2(y-x).

log(y^2/2x) = log2y-log2x. this is where I think the divison must apply to the right hand ones.
(y/2x) =(2y-2x) Taking antilogs

(y^2/2x) = 2x(2y-2x)

y^2 = 4xy-4x^2

y^2-4xy+4x^2 = 0

y^2-2xy-2xy+4x^2=0

y(y-2x)-2x(y-2x)=0

(y-2x)(y-2x)=0

(y-2x)^2.

My question is why not log2y-2x is not made to be log(2y/2x), since it has that negative sign? In order to go by the  law.