# Simplification of Powers of Five

#### CSmith

##### Member
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

#### SuperSonic4

##### Well-known member
MHB Math Helper
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

#### Prove It

##### Well-known member
MHB Math Helper
1. r = 0 (anything to the 0th power is 1)
Except $\displaystyle 0^0$ which is undefined...

#### chisigma

##### Well-known member
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

$\chi$ $\sigma$