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Then that would be decimal numbers I think.
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That is correct. I am going to eat see you in a half hour.
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Now before I proceed on the decimals, let me endeavour to put this across c'os I am getting zero as an answer to this and the book seems to have 1/4 ;
logx - 4logx = 2
The LHS is having 8 as their bases
Thank you!
Last edited by EbenezerSon (2014-09-28 04:49:05)
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Hi;
I am getting 1 / 4 as well.
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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Is 8 not going to raise the power 2 at the RHS?
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First algebraically clean it up a bit.
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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Like this?
x - x^4 = 8^2
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logx - 4logx = 2
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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logx - 4logx = 2
wierd to me though
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If you had 1 apple and you took 4 apples away you would have -3 apples. 1 log(x) - 4 log(x) = 3 log(x). 1 of anything - 4 of the same thing is -3 of that thing.
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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How I did it was;
log(x/x^4) = 8^2
taking antilog
x = 64x^4
[I was rather thinking that minus sign in logarithm means a division sign, what do you say to this?]
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That is not correct. When you take the anti-log, you have to take it of both sides.
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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logx - 4logx = 2
I'm now grasping it - how are you gonna proceed from here?
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Divide both sides by -3.
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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I see now, but do you agree that a minus sign in logarithm means one has to divide?
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Yes, that is the reason tjhey were invented for.
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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If so, then I was thinking the steps should be;
logx - 4logx = 2
log(x/x^4) = log8^2
taking antilog of both sides
x/x^4 = 8^2
x = 64x^4
What do you say to this?
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It is not correct. Where does log8^2 come 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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Where does log8^2 come from?
That's according to the difinition of logarithm
- "the logarithm of any number b to the base a is the index or power to which the base must be raised to give the number" example ;
log b = c [taking 'a' as the base] = b = a^c
Last edited by EbenezerSon (2014-10-01 04:57:57)
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That is written
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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There was a similar problem in the book, and there was 1 at the right hand side of the equation, and in the course of manipulation it became [log10] and when antilog was taken it became 10. So that's the idea I tried using it. I don't know if you understand what I am putting across.
Hi; this is the problem I spoke of above, that used the one I understand;
log(x^2 + 1 ) - 2logx = 1
log(x^ + 1/ x^2) = log10
x^ + 1/x^2 = 10
x^2 + 1 = 10x^2
9x^2 = 1
x^2 = 1/9
x = +1/3
Please, see the method - this is the one I wanted to use for the one under discussion
Last edited by EbenezerSon (2014-10-17 01:19:26)
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I understand but it leads to the wrong answer, therefore I would drop it.
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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Okay, I will let you see that problem yourself.
In one of the rules of logarithm it says, log(m/n) equal log m - log n, so I was expecting you would say logx - 4logx would be log(x/x^4). But couldn't understand why you subtracted logx from 4logx.
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There is nothing wrong with your idea except that because it is complicated you are making some mistake on the next line.
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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Please, if you would not mind use my idea to solve, because I am rather familiar with that idea.
Thank you very much
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