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

I think 5 is raising to the power nothing therefore should be equal to zero and not one.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

Supposing you had

2^4 * 2^0

when we multiply the same bases we are allowed to add the exponents. 4 + 0 = 4 so 2^4 * 2^0 = 2^4 so 2^0 must equal 1.

**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 is it possible to add the exponents of the following?

*Last edited by EbenezerSon (2013-07-23 23:45:13)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

No, it is not. They do not have the same base.

**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 is due to such situations why I could not solve the previous problem since it has 6*3^2n+3. the six made it hard for me to manipulate. One cannot reduce it to be three. Or if the six had been three I would multiply it with the three to get nine and then reduce it to three, so I could manipulate. Since nine would be a perfect square.

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

Then the final answer would be 9*16 = 144, is that right?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

6 = 2 * 3 so you could have combined that statement into

6*3^(2n+3) = 2 *3 * 3^(2n+3)= 2 * 3^(2n+4)

and you are done.

**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:

6 = 2 * 3 so you could have combined that statement into

6*3^(2n+3) = 2 *3 * 3^(2n+3)= 2 * 3^(2n+4)

and you are done.

I don't understand yours, I thought it would be

2*3*3(2n+3) = 2*9^(2n+3) = 2*3^3(2n+3) = 2*3^(6n+9)

*Last edited by EbenezerSon (2013-07-24 00:20:56)*

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

I must review your procedure with the previous problem to comprehend.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

You add coefficients:

2 * 3^1 * 3^(2n+3)

2n +3 + 1 = 2n + 4 so

2 * 3^(2n+4)

Are you saying that

because that 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

Errrr okay I have got you, then what would one make of the two, I mean the base.

*Last edited by EbenezerSon (2013-07-24 00:33:59)*

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

bobbym wrote:

You add coefficients:

2 * 3^1 * 3^(2n+3)

2n +3 + 1 = 2n + 4 so

2 * 3^(2n+4)

Are you saying that

because that is incorrect.

The two I have underlined above.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

Hi;

You can only do this:

6 * 3^(2n+3)

2 * 3 * 3^(2n+3)

2 * 3^1 * 3^(2n+3)

Add the exponents on the threes.

2n +3 + 1 = 2n + 4 so

2 * 3^(2n+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

Okay I digest it. But can I use that procedure to arrive on that answer 7 you had in the previous problem?

*Last edited by EbenezerSon (2013-07-24 00:44:58)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

That problem is a real bear and I recommend the other method, the one I posted.

I am going to get a little bit of sleep see you later.

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, Thanks God bless. I shall solve more problems.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

How did you do?

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

Hi, bobbym I have come across another confusing one with different bases.

Solve for X.

3^x * 2^x-1 = 1.

How would one go about this? Is wierd.

*Last edited by EbenezerSon (2013-07-24 04:01:51)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

Is that

because if it is...

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

The book has 0.39 answer

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

That is correct if you round my answer to 2 decimal places.

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:

Is that

because if it is...

Please help me understand this.

Thanks in advance

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

I can show you the steps but unless you have a little knowledge about logarithms it will be confusing.

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 had solved much problems on logarithms.

Usually I had encountered problems on indices with unequal bases, and the book used log to solve but I can't percieve this would need logarithm application.

Please carry on!

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,388

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.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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