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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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?

Please assist.

*Last edited by EbenezerSon (2013-08-03 10:00:14)*

I know only one thing - that is that I know nothing

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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?

I know only one thing - that is that I know nothing

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

Which textbook are these questions from?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

I have not asked him that yet.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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

Thanks.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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.

I know only one thing - that is that I know nothing

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

p=1000.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

1/3logp=1

Times both sides by 3

log(p) = 3

Take 10^ of both sides.

p = 10^3 = 1000

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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.

I know only one thing - that is that I know nothing

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Line 3 is incorrect.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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?

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

First one is okay.

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

You mean cube root.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

yes, yes, cube root instead.

But are they all right in that manner?

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Yes, they are correct.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Have you sorted out the x and y problem?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

What x and y problem?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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.

Here lies the reader who will never open this book. He is forever dead.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

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**.

Please clear my doubt.

Thanks much.

*Last edited by EbenezerSon (2013-08-04 10:14:08)*

I know only one thing - that is that I know nothing

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