MHB Simplification of Powers of Five

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The discussion focuses on expressing various values in the form of 5^r, where r is a rational number. Participants correctly identify the values of r for several expressions, such as r=0 for 1, r=-1 for 1/5, r=7/2 for (sqrt(5))^7, r=2/3 for the cube root of 25, r=3/2 for the square root of 125, and r=-1/2 for 1/sqrt(5). There is an emphasis on the importance of presenting answers in the specified format, as examiners may deduct points for not doing so. The conversation also touches on the nuances of exponentiation and the definition of sequences related to powers.
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write in the form 5^r, where r is a rational number.

1.)1
2.)1/5
3)(square root 5)^7
4.)3 suare root 25
5.)square root 125
6.)1/ square roor 5



1. r = 0 (anything to the 0th power is 1)

2. y = 5^x
1/5 = 5^x
x = -1

3. sqrt(5)^7 = 5^x
x = 7/2
4. cube root (25) = 5^x --->
25 ^1/3 = 5^x
25= 5^2
x = 2/3

5. sqrt(125) = 5^x
125 = 5^3
125^1/2 = 5^x
5^3/2 = 5^x
x = 3/2

6. 1/sqrt(5) = 5^x
5^-1/2 = 5^x
x = -1/2
 
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CSmith said:
write in the form 5^r, where r is a rational number.

1.)1
2.)1/5
3)(square root 5)^7
4.)3 suare root 25
5.)square root 125
6.)1/ square roor 5



1. r = 0 (anything to the 0th power is 1)

2. y = 5^x
1/5 = 5^x
x = -1

3. sqrt(5)^7 = 5^x
x = 7/2
4. cube root (25) = 5^x --->
25 ^1/3 = 5^x
25= 5^2
x = 2/3

5. sqrt(125) = 5^x
125 = 5^3
125^1/2 = 5^x
5^3/2 = 5^x
x = 3/2

6. 1/sqrt(5) = 5^x
5^-1/2 = 5^x
x = -1/2

Mostly correct (you have the value of all your exponents correct). What you need to watch out for is that the question states write in the form 5^r... meaning your answers should be written as $5^r$ instead of working out r and leaving it there.
If I take your first one as an example you correctly figured out that r=0 so you'd put $5^0$ as your answer instead of 0.

I know it seems trivial (and it probably is) but that's the sort of thing examiners love taking away an answer mark for (Devil)
 
CSmith said:
1. r = 0 (anything to the 0th power is 1)

Except $ \displaystyle 0^0 $ which is undefined...
 
CSmith said:
1. r = 0 (anything to the 0th power is 1)

What CSmith has written is correct even if some care has to be used in order to avoid confusion. As explained in... http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

... the exponential sequence on base $\displaystyle\alpha$ the general term of which is usually written as $\displaystyle\alpha^{n}$ is defined by the recursive relation…

$\displaystyle a_{n+1}= \alpha\ a_{n}\ ,\ a_{0}=1\ ,\ \alpha \in \mathbb{R}$ (1)

... so that the sequence $\displaystyle \alpha^{n}$ is $1\ ,\ \alpha\ ,\ \alpha^{2}\ ,\ ...$. That is true of course also for $\displaystyle \alpha=0$ so that the sequence $\displaystyle 0^{n}$ is $1\ ,\ 0\ ,\ 0\ ,\ ...$.

In a fully different situation we are when if we have to valuate the limit...

$\displaystyle \lim_{x \rightarrow 0} \alpha(x)^ {\gamma(x)}$ (2)

... where $\alpha(0)=\gamma(0)=0$. Here the term indeterminate form is correctly used because the limit (2) is not 'automatically' equal to 1 in any case...

Kind regards

$\chi$ $\sigma$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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