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

Now I have learnt that! Thanks much Bobbym!

Okay, back to my question.

27^(n+2) =27^n * 27^2 = 3^(3n) * 3^(6)

.

Why didn't you use 3^(3n) * 3^(6) but rather wrote 9^(n+2) * 3^(n+2).

Thanks in advance.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

All that is correct.

Why didn't you use 3^(3n) * 3^(6) but rather wrote 9^(n+2) * 3^(n+2).

Because it was easier to cancel.

**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

Candidly speaking, since I started working on indices I haven't seen nor come accros a problem that could produce different bases as your own. As you made 9^(n+2) *3^(n+2) out of 27^(n+2)

Now I think the method that will be applicable to a problem is the one that must be used.

Thank very much Bobbym, God bless you.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

Hi;

Glad to help and let me know when you need more.

**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

bobbym wrote:

Okay, I will provide the steps:

Take the log of both sides.

Add log(2) to both sides.

Divide both sides by ( log(2) + log(3) ).

And we are done.

Hi

Initially there was no addition sign[+] but in the course of manipulation you happen to bring an addition sign instead of the multiplication sign.

Please, explain why

thanks

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

Hi;

Line 2?

**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

Besides, there was two 'x' as in xlog3 + xlog2, And you didn't add those two x to be '2x', or in other words, why didn't you add those x's to be 2x after you divided log2 by log3 + log2, do you understand my question?

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

bobbym wrote:

Hi;

Line 2?

yes

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

The general rule:

Log(ab) = Log(a) + Log(b)

Under certain conditions of course.

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 answer my question at #282, if it is unintelligible please tell me

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

Besides, there was two 'x' as in xlog3 + xlog2, And you didn't add those two x to be '2x', or in other words, why didn't you add those x's to be 2x after you divided log2 by log3 + log2, do you understand my question?

Please, what line do you mean?

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 line 4, there is one 'x' behind log2 and another 'x' behind log3, why didn't you add those xs so that it would have been 2x.

Please check these for me:

x( y^1/2 - x^1/2)

= (xy^3/2 - x^3/2)

Is my answer correct?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

Hi;

I am afraid that

x log(2) + x log(3) does not allow us to add the x's and say 2x.

x( y^1/2 - x^1/2)

= (xy^3/2 - x^3/2)

Is my answer correct?

That is not correct, please try again.

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 think you want me to say; (xy^1/2 - x^1/2)

But I have in mind that the 'x'(the x which is outside of bracket) is raising to the power 1, but the 1 is not clearly written and that I am thinking my answer at 287 is correct, what do you say?

*Last edited by EbenezerSon (2014-08-24 04:17:39)*

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

Hi;

I think you want me to say; (xy^1/2 - x^1/2)

Not exactly, that is incorrect also. You are correct in stating that x is x^1. Now just go through it piece by piece adding the exponents of x.

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 have done that, still I only arrive at the same answer.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

Hi;

What is x( y^1/2 )?

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

To begin with I would say the X instrinsically raised to the power positive one[1], so I will say 1+1/2 = 3/2. Therefore;

XY^3/2

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

But there is no x in there. x^1 = x. What you have is x^0 = 1.

So x( y^1/2 ) = x y^1/2

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 was talking of the X which is outside the bracket. Look at a similar case at #182 line one, after you had multiplied 2^-x through, the 2 at right side became 2^x+1, this indicates that the 2 already had 1 as its power, and in this problem I perceive that the X has 1 as its power though not visibly expressed or written , what do you say?

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: 86,443

In the parentheses of x( y^1/2 ), there is only a y to the (1/2) power. There is no x in there other than x^0 which equals 1 so we do not include it.

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

So, if there was X in the bracket having 1/2 as it power how would the answer be? like x(y^1/2 - x^1/2)

Thank you.

*Last edited by EbenezerSon (2014-08-27 04:00:48)*

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

For x(y^1/2 - x^1/2)

you do the first term first:

then the second term:

because you add the exponents which are 1 and 1 / 2 = 3 / 2.

Then you finish off by combining the two terms.

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

Why did you not say x(y^1/2) = xy^3/2 I think the x has 1 as its power, please I am not getting you with this x(y^1/2) = xy^1/2. Please explicate.

*Last edited by EbenezerSon (2014-08-27 04:49:50)*

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,443

Hi;

You can only add the exponents when the bases are the same:

both are x's so add the exponents. both are y's so add the exponents. , it just stays the same. You can do nothing here because the x and y are different.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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