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## #151 2013-07-24 21:50:40

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

### Re: Simplify the following:

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.

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

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## #152 2013-07-24 21:53:15

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

Hi;

This step?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #153 2013-07-24 22:04:51

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

### Re: Simplify the following:

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

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

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## #154 2013-07-24 22:09:50

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

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.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #155 2013-07-24 22:26:59

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

### Re: Simplify the following:

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.

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

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## #156 2013-07-24 22:29:12

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

What example do you mean?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #157 2013-07-24 22:55:13

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

### Re: Simplify the following:

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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## #158 2013-07-24 22:57:25

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

Hi;

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

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #159 2013-07-24 23:14:59

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

### Re: Simplify the following:

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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## #160 2013-07-24 23:19:53

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

2 - 2 log(5) is the same as 2 - log(5^2)  you can not switch them.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #161 2013-07-25 00:16:17

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

### Re: Simplify the following:

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.
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,

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

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

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## #162 2013-07-25 00:33:26

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

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

Where are you getting that 10 from?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #163 2013-07-25 00:41:33

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

### Re: Simplify the following:

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?

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

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## #164 2013-07-25 00:45:10

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

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

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #165 2013-07-25 00:59:17

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

### Re: Simplify the following:

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

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## #166 2013-07-25 01:02:51

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

(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

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #167 2013-07-25 01:15:07

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

### Re: Simplify the following:

And why log10^2 in # 163 correct?

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

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## #168 2013-07-25 01:17:39

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

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

http://www.mathsisfun.com/definitions/d … e-law.html

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #169 2013-07-25 03:58:41

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

### Re: Simplify the following:

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.

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

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## #170 2013-07-25 15:21:48

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

What law says that?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #171 2013-07-25 19:32:59

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

### Re: Simplify the following:

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?

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

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## #172 2013-07-25 20:03:09

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

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.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #173 2013-07-26 05:32:47

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

### Re: Simplify the following:

bobbym wrote:

Okay, I will provide the steps:

Take the log of 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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## #174 2013-07-26 12:46:46

bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 107,123

### Re: Simplify the following:

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.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #175 2013-07-27 03:54:01

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

### Re: Simplify the following:

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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