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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Hi;

I did that to get x all by itself on the left. What happened is I subtracted log(2) from both sides. That moved it over to the right side leaving a simpler expression on the left.

In case you are having some problems with balancing an equation check this out.

**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

Thank you very much bobbym for the link.

But there was xlog3 also present at the left side with the log2. Why didn't you balance the eqution with the log3 instead? because it was also present at the left side.

I just want to why you didnt use the log3 instead.

Thanks.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Hi;

If I were to move the x log(3) to the right side I would have to subtract x log(3) from both sides. Now I would have x on both sides. Remember I am trying to get x all byt itself and on one side.

**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

Okay, but bobbym, the law logx-logy= logx/logy.

If it had been, logy-logy=0, is the zero?

If the logs have the same figures, it not right for one to divide?

But rather subtraction must take place?

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

You had your final answer by this:

x = log2/log2+log3

= 1.477

But I cant tell why the has, 0.39

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

Please, see how I solved the following:

2^x +2^-x=2.

bases are the same, therefore equate the exponents.

x-1-x =1

= x-x-1=1

=0-1 =1

=0 = 1+1

=0=2.

Is that right?

I think bases are the same, so doesnt need logarithm application.

*Last edited by EbenezerSon (2013-07-27 22:35:16)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Hi;

What answer did you get for x?

I would do it like this:

Multiple everything by 2^x to clear fractions out.

Factor the left side.

So

(2x-1) = 0 and (2x-1) =0

2x-1 = 0 then x = 0

2x-1 = 0 then x = 0

x = 0 is the answer.

**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

I had zero as the answer.

But as I am looking through the book, I chanced on the following, which I have solved but one of the answers was impossible to get. And the book had zero with that answer.

3^x + 3^-1=4.

My final answers are, x=1, and the impossible answer, 3^x=4.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Hi;

I am afraid x = 1 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

The book has x=0 or 1 as answer at its back.

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

How then would one solve this?

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

Okay, I admit, then see my thoughful manipulation:

3^x + 3^x-1= 4

multiplying 3 through.

3*3^x + 3^x*1/3 = 4* 3.

Let 3^x=m. therefore,

=3m + m = 12

= 4m - 12 =

0

factorising.

= 4(m-3) =0.

Is this correct?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Hi;

What did you get for 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

The final result is my answer

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

I have considered it carefully, but see no way to solve than what I have done, please assist.

Thanks.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Hi;

Factor out a 3^(x-1) from each term on the left.

Divide both sides by 4.

Take log of both sides, you get.

(x-1) log(3) = 0

Divide both sides by log(3)

x - 1 = 0

x = 1

See you later, I need to do some chores and shopping.

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:

Hi;

Factor out a 3^(x-1) from each term on the left.

I don't understand how you factored it out.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Hi;

3* 3^(x-1) = 3^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

Please check, the following for me.

Log(2x+3)=log4.

I had 20 as the answer, the has 10.

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.

log(10+9x)-log(11-x)=2. I had 20 as answer, the book has 10.

Last but not the least

log81/log3^1.

the book has -4 at its back as the answer.

Thanks.

Please, check the above if the answers are right. Me vs the book. -:)

*Last edited by EbenezerSon (2013-08-03 07:20:08)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Hi;

Log(2x+3)=log4.

I had 20 as the answer, the has 10.

You had what for 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 had 20.

The book has 10 as the answer

Sorry I supposed to have checked the writings before posting.

Thanks.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

Both answers are wrong. The correct answer is 1.

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

how about if the log4 were to be log45. What would the answer be?

Thanks.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,509

You mean log(45) then x = 21.

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

thanks, but please see the rest if they are right with their respective answers from the book, and my answers as well.

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