#1**+2 **

First one.....convert to exponential form

5 ( 4n)^(1/4)

__________

2 (27n)^(1/4)

We can "cancel" the n's and get

5 (4)^(1/4)

_________ multiply top/bottom by (27)^(3/4)

2 (27)^(1/4)

5 ( 4 )^(1/4) (27)^(3/4)

__________________

2 (27)^(1/4) 27^(3/4)

5 [ 4 (27)^3 ] ^(1/4)

_________________

2 * 27

5 [ 4 (3^3)^3 ] ^(1/4)

________________

54

5 [ 4 * 3^9 ]^(1/4)

______________

54

Note : (3^9)^(1/4) = 3^(9/4) = 3^(2 + 1/4)...we can take 3^2 out of the radical and leave one power of 3 inside

5 * 3^2 [ 4 * 3 ]^(1/4)

_________________

54

5 * 9 [ 12]^(1/4)

_____________

54

5 [12]^(1/4)

_________

6

5 ^{4}√12

______

6

CPhill
Oct 24, 2018

#2**0 **

Thanks for helping! I still don't quite understand though, there are a lot of steps and it's confusing >.<

RainbowPanda
Oct 24, 2018

#3**+2 **

Yeah I know.....we are using a lot of exponential rules here

Here are a few that I employed

^{m} √ [ a^n ] converts to exponential form (a)^(n/m)

And

a^(m) * a^(n) = a^(m + n)

And

Remember.... we can take something out of a radical when we have (a)^(n/m) and n ≥ m

For instance...let's suppose that we have ^{ 3} √ [ 4^5]

In exponential form....we have (4)^(5/3)

To determine how many powers of 4 come out of the radical and how many stay in.....write the exponent as a mixed number

4^(1 +2/3)

The whole number tells us that 1 power of 4 comes out of the radical and the numerator of the fraction tells us that 2 powers of 4 stay inside the radical...so we have

4 ^{3}√ [4^2 ] = 4 ^{3}√ [ 16 ]

But note that 16 = 2^4....so...we can apply this rule once more

We have 4 ^{3}√ [2^4 ] = 4 (2)^(4/3)

To determine how many powers of 2 that we can take out of the radical and how many we can leave in...write this as

4 (2)^(1 + 1/3)....so....we can take 1 power of 2 out of the radical and 1 power of 2 stays in

So..putting this all together, we have

(4 * 2) ^{3}√2 = 8 ^{3}√2

Hang in there, RP......you will begin to understand this

CPhill
Oct 24, 2018

#6**+2 **

The second one is easier, RP

√ [5 p^3 ]

_________ = (1/3) √ [ (5 p^3) / (2p^3) ]

3 √ [2 p^3 ]

The p^3's will "cancel" and we are left with

(1/3) √ [ 5 / 2 ] =

√ 5

_____ (do you see this ??? )

3 √2

We want to get rid of the radical in the denominator [ this is called "rationalizing the denominator" ]

We can do this in this manner.....Multiply top/bottom by √2 and note that √2 *√2 = √4 = 2

√5 * √2

__________

3 √2 * √2

√10

_____

3 * 2

√10

_____

6

CPhill
Oct 24, 2018

#8**0 **

I let you in on a secret, RP....

In most of the higher math classes I've had....I've rarely encountered these types of radicals.....these are mostly " textbook" exercises.....once you have to suffer through this.....you probably won't see this stuff, again !!!!!

[ Something to look forward to, huh ??? ]

CPhill
Oct 24, 2018