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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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: 103,709

Hi;

This step?

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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
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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.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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: 103,709

What example do you mean?

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,709

Hi;

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

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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.

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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.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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: 103,709

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

Where are you getting that 10 from?

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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: 103,709

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

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

Please I what you to answer my question at #161

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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(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

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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: 103,709

log(10^2) is 2 log(10). log(10) in base 10 is 1.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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
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What law says that?

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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: 103,709

That is not the law:

log2-log2 = log2/log2.

Take a close look at that,

log2-log2 = 0

log2/log2. = 1

0 ≠ 1

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 510

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.

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