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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,525

Something is wrong there. Are you sure you copied the problem correctly?

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: 86,676

log(y^2/2x) = log2y-log2x

The right hand side of that is incorrect.

**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

it's log2(y-x)

I multiplied log2 with what is in the bracket, to have that. Why you say so.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi;

Because when you write

log2y-log2x

how do you know which of these is what you want

log(2)*y or log(2y)?

**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

Errr, I dont normally put bracket around figures as you normally do.

But I suppose each of them are the same, I mean what yours above.

In #229

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Errr, I dont normally put bracket around figures as you normally do.

Math has to be precise. More precise than anything else you do.

When you write log2x, no one can figure out what you mean. It appears you wanted log(2)x, but I am not sure. Which do you want?

**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

Thanks, Bobbym.

Then please let me how it should be

By the way I have developed the attitude of not putting brackets c'os the book I have doesn't put it around each example it gives. And that has become of me.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi;

In post #225 you wrote this

log(y^2/2x) = log2y-log2x

Please use brackets to tell me what you want on the right hand side.

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 Bobbym I havent envisioned that bracket of the right hand side could cause any thing, but I think;

Log2(y-x) = log2y-log2x

I think is the same as, log(2y)-log(2x)

Please help if I am confused.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Okay, we have the original problem looking like this

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.

Hold on while I work on it. The book answer is okay but it is not an answer to the question

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

Oops. Bobym.

Please I made a little blunder!

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

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

So you want this?

It looks a lot like the earlier one.

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

2logy-log2x=log2(y-x)

express y in terms of x.

This how is in the book.

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: 86,676

That book is awful.

Mathematics uses brackets to avoid ambiguities.

I will interpret it as:

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 is the answer I had the same as yours when you solved?

I am using my phone to browse and the battery has run low.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Those are all correct answers with methods shown.

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

Is your interpretation the ones the book should have used?

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Let's see what happens.

10^ to both sides

Multiply both sides by 2x

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

anonimnystefy wrote:

Hi EbenezerSon

With a careful look at signs I think should be,

x-3x-3-3/x^2-9 =

-2x-6/x^2-3^2 =

-2(x+3)/(x-3)(x+3)

= -2/x-3

Please am I right there.

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: 86,676

I am getting

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

Oops,

I have considered it once more and I could see Anonymstify is right with his answer, I lost sight with some negative sign, here

x+3-3(x-3) = x+3-3x+3.

=-2x+6

Bobbym, I suppose you lost sight on a negative sign. With a careful check you will adentify it.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

Hi;

I am sorry but my answer is correct.

Please check post #244, in line 2 there is an error.

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

Please could you demostrate to me?, c'os I can't adentify it.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,676

How did 3(x+1) become 3x-3?

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

I see, then let me check the original one and come back.

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

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