Solving with indices
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Solving with indicesBelow is a question which I need help on:
Solve the following equation:
I do understand that 81 must be set as x to the power of y. Also I understand that the 32 must be set to the same value of x (same) to the power of z (different):
How do I get the value for x??
How do I know the value for x??
Thanks a lot for any help provided. -
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Re: Solving with indices16 -->32 is denominater (first fraction) -->numerator (second fraction)(Original post by james22)
Use logs?
If you cannot use logs for any reason then notice that n must be negative. How do you get from 16 to 32 using powers? (both are powers of 2). Same question for 81 to 243? Can you guess at a value for n now?
Are you really allowed to do this????????? Or are you doing it becasuse you somehow know that n is negative?
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Re: Solving with indicesBut the numerator of the first fraction is not equal to the numerator of the second fraction.(Original post by notnek)
It may make it simpler for you if you write each number as a product of primes:
This should give you a clue as to why the power is negative and lead you to the solution. How do I deal with this problem?
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Re: Solving with indicesso try flipping them and adjusting the power accordingly...(Original post by krisshP)
But the numerator of the first fraction is not equal to the numerator of the second fraction. How do I deal with this problem?
Last edited by Plato's Trousers; 06-07-2012 at 10:48. -
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Re: Solving with indicessee my post #9 - get it now?(Original post by krisshP)
btw if a power is negative, don't you usually put the number as a fraction with 1 as the numerator? -
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Re: Solving with indicesI know n is negative because you cannot (easily) get from the numerator of the first to the numerator of the second fraction. However this is much easier if the first fraction is flipped. If n is negative the fraction is flipped so i assumed that n is negative because it makes things easier.(Original post by krisshP)
16 -->32 is denominater (first fraction) -->numerator (second fraction)
Are you really allowed to do this????????? Or are you doing it becasuse you somehow know that n is negative?
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Re: Solving with indicesit's better not to use decimals for indices. use(Original post by krisshP)
ok. I got the correct answer of n=-1.25.
Thanks a lot for making me learn
Your help is much appreciated.
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Re: Solving with indicesIt's just easier to see what's going on if you use a fractional index. The top number is the power and the bottom number is the root.
So, for example
![x^{\frac{5}{4}}=\sqrt[4]{x^5}](http://www.thestudentroom.co.uk/latexrender/pictures/4a/4a5d7a0b8ec7b658ca075f9d7fa8649d.png "x^{\frac{5}{4}}=\sqrt[4]{x^5}")
and a minus sign indicates a reciprocal, so
![x^{-\frac{5}{4}}=\dfrac{1}{\sqrt[4]{x^5}}](http://www.thestudentroom.co.uk/latexrender/pictures/af/afa0764462ff9220337a79db8f5bfe5f.png "x^{-\frac{5}{4}}=\dfrac{1}{\sqrt[4]{x^5}}")
Last edited by Plato's Trousers; 06-07-2012 at 10:59. -
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Re: Solving with indicesYou probably already got this by now, but just to clarify:(Original post by krisshP)
btw if a power is negative, don't you usually put the number as a fraction with 1 as the numerator?
The degenerate case is when b=1:
Alternatively, we can start by assuming the degenerate case to be true (since you accepted that):
Thus, as the degenerate case holds true:
