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last 2

RainbowPanda  Oct 24, 2018
 #1
avatar+90968 
+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

 

 

 

cool cool cool

CPhill  Oct 24, 2018
 #2
avatar+2051 
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
avatar+90968 
+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

 

 

cool cool cool

CPhill  Oct 24, 2018
 #4
avatar+2051 
+1

That is a lot to process...

RainbowPanda  Oct 24, 2018
 #5
avatar+90968 
+1

Agreed........I'll try to answer your second problem with extensive explanations

 

 

cool cool cool

CPhill  Oct 24, 2018
 #6
avatar+90968 
+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

 

 

cool cool cool

CPhill  Oct 24, 2018
 #7
avatar+2051 
+1

Yeah I got that one, it was easier than the other one

RainbowPanda  Oct 24, 2018
 #8
avatar+90968 
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 ???  ]

 

 

 

cool cool cool

CPhill  Oct 24, 2018

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