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

I don't seem to understand the step 3.

I thought, it would be something like:

xlog10 - log2 + xlog3 = 0

I claim the above because of the following:

2 - 2log5 = log10^2 - log5^2 = log4.

Why didnt it take xlog10. because 2 is the log and not the base. I think it cant be xlog2.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

Hi;

This step?

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

Yes. if you look at the my example, log ten raises to the power two. And not log5^2

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

log(3^x) breaks down into xlog(3) regarldless of base 10 or base e or base 2.

Same for log(2^(x-1)) = (x-1) log(2). According to the rule

log(a^b) = b * log(a)

**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

Yes, I understand that law, so now, log2^x-1= xlog2 - log2

So does my example mean, 2 - 2log5 = log5^2-2.

If so why is it equall to log10^2 - log5^2. and not log5^2 - log5^2 in order to go according to the law.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

What example do you mean?

**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

The book has this example.

2 - 2log5 = log10 - log5^2 =log4

why should not it be, 2 - 2log5 = "log5^2" -log5^2 so that it will go by the law.

With example it has log10^2, I think should be log5^2

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

Hi;

You are done with the old problem and wish to do another 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

I am comparing two things because I dont understand your step 2.

The is an example from the book: 2-2log5 = log10^2 - log5^2 = log4

so does 2-2log5 mean log5^2-2?

That what I want to know.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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2 - 2 log(5) is the same as 2 - log(5^2) you can not switch 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

How different is this x-1log2 in the previous problem from 2-2log5 in the example.

Because I see it that one must apply the same method, but the method used to solve both are different.

Please see this;

x-1 log2 = xlog2-1log2.

(The same base used).

And

2-2log5 = log10^2-log5^2.

(Different bases used).

why is it log10^2 and not log5^2 instead. So that the whole thing would be;

log5^2-log5^2,

Please, explain to me why.

*Last edited by EbenezerSon (2013-07-25 00:33:48)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

2-2log5 = log10^2-log5^2

Where are you getting that 10 from?

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

Thats what is written in the book.

It used log ten for the unknown log of the two. Hence

2-2log5=log10^2-log5^2

But I dont know why.

Is that wrong if so how should it be?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

No, that is correct. I did not see it at first.

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 I what you to answer my question at #161

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

(x-1) log2 = xlog2-1log2.

That has nothing to do with bases of exponents. It uses another law of algebra called the distributive property:

a(b + c) = ab + ac or (b + c)a = ab + ac

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 provide a link for that.

Is not among the law of log in my book.

And why log10^2 in # 163 correct?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

log(10^2) is 2 log(10). log(10) in base 10 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

bobbym wrote:

Okay, I will provide the steps:

Take the log of both sides.

Why the last step is not:

Since the law says. for instance log2-log2 = log2/log2.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

What law says that?

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

This is what I mean, and it is one of the laws of logarithm in my book:

logx - logy = log(x/y).

This is what I mean, and I see you did not apply it at that step. Why didn't you?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

That is not the law:

log2-log2 = log2/log2.

Take a close look at that,

log2-log2 = 0

log2/log2. = 1

0 ≠ 1

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

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**

**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.

.

Then how did you arrived at the last stage? Could you explain to me?

Why did you added log two to both sides?

Thanks

*Last edited by EbenezerSon (2013-07-26 05:34:29)*

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,319

If you were solving

x - 2 = 0

you would add 2 to both sides

x -2 + 2 = 2

-2 + 2 = 0

x = 2.

Notice how x is all by itself on the left side.

I am doing the same thing in that equation. Trying to get x all by itself on the left.

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 I am sure if you had not added log2 to both sides you would still arrive at the same answer, is not that?

------------------

xlog2 - "log2" +xlog3 =0. From this stage you took off the "log2", why that happened.

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

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