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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,918

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

The right hand side of that is incorrect.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

it's log2(y-x)

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

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.

**In mathematics, you don't understand things. You just get used to them.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

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

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

Oops. Bobym.

Please I made a little blunder!

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

So you want this?

It looks a lot like the earlier one.

**In mathematics, you don't understand things. You just get used to them.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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: 93,747

That book is awful.

Mathematics uses brackets to avoid ambiguities.

I will interpret it as:

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

Those are all correct answers with methods shown.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

Is your interpretation the ones the book should have used?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

Let's see what happens.

10^ to both sides

Multiply both sides by 2x

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

I am getting

**In mathematics, you don't understand things. You just get used to them.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

Hi;

I am sorry but my answer is correct.

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

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,747

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

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 476

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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