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First, remember that:
edited typo mistake... Thanks RickyWe need to figure out what to multiply both sides of your first equation. We want to get the exponents equal:
So multiply both sides of the first equation by 2 raised to the 2x - 1 power:
Last edited by pi man (2006-11-02 09:02:40)
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You could also do this:
Now substitute this into the second part:
This has the advantage of only depending on y.
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pi man
thank U very much,,,
polylog
yessssss,,
that's good work.
thank you veeeery much..
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First, remember that:
This is incorrect.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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i know that people have already gave solution, but theres my workings out of it
The Beginning Of All Things To End.
The End Of All Things To Come.
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hmmmmmmmmmm,,
What multipied by 2^x-1 equals 2^3x-2 ?.
(2^x-1)(2^2x-1)=2^3x-2
2(2^x-1)^3=2^3x-2
= 2y^3
Is that a correct answer ?
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that is a correct answer, theres nothing wrong in the working out. I just assumed you were meant to use logarithms for this, which really if you know how they work, is a simpler method than yours, since using logarithms requires no real thinking at all. wheras yours requires you to think of something that multiply the first statement to give the second, and then thinking of how to rearrange it to keep it the same, but continaing the original statement in a form you can reduce into an equation containing y, if you know what i mean.
The Beginning Of All Things To End.
The End Of All Things To Come.
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thank you ALL
:D:cool::cool:
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