Question

How do you solve this problem? Please explain!

Answer 1

\[(4a ^{5/3})^{3/2}\]

Answer 2

Solve for what?

Answer 3

It says to simplify in this problem

Answer 4

basic rule is \(\large (b^n)^m = b^{n \times m}\) because it's \(b^n \times b^n \times b^n ......\) m times. so \(\large (a. \ b^n)^m = a^m \ . \ b^{n \times m}\)

But you can go at this in a number of ways; and you might wish to try this approach:

\(\large (4a^{5/3})^{3/2} = \sqrt{4a^{5/3} \ * 4a^{5/3} \ * 4a^{5/3} }\)

it's not the immediate solution, you will get that by looking up the exponent rules as posted above,

but it might help get you get comfortable with these kind of questions.....maybe

under the \(^{3/2}\) umbrella you are cubing and square rooting....

Answer 5

How would I multiply the numbers inside the square root?

Answer 6

well, imagine the square root is not there. jut multiply those out as you normally would:

\(4 a^{5/3} \times 4 a^{5/3} \times4 a^{5/3} \)

we'll worry about the square root sign after we have done that.

Answer 7

Okay. When you are multiplying exponents you have to add them right?

Answer 8

when multiplying the same base , you add the exponents.

Answer 9

So without the square root it equals 4a^5?

Answer 10

if you remember that x*x*x is x^3
but x is x^1 and x*x is x^2
you get the example
x^1 * x^2 = x^3 (that is how I remember this rule)

Answer 11

yes, but you also have 4*4*4 to do

Answer 12

nah

you also have 4 x 4 x 4!!

Answer 13

64a^5?

Answer 14

fine

i'll leave it to @phi

you are in great hands!! good luck

Answer 15

then it would equal 8a.....how would I square root the 5?

Answer 16

a^5 is a*a*a*a * a
we can do square root of a^4 (it is a*a or a^2)
square root of the "extra" a is either sqr(a) or a^1/2

Answer 17

\[ \sqrt{64 a^4 } \sqrt{a} \]

Answer 18

So the answer is \[8a \sqrt{a}\]?

Answer 19

8a^2 sqr(a)

Answer 20

So that would be simplified?

Answer 21

they probably want you to use exponents, so write sqr(a) as a^(1/2)

Answer 22

\[8a ^{2}a ^{1/2}\]

Answer 23

Like that?

Answer 24

yes, and though it looks a bit weird, we are multiplying the same base. (the a)
so we add the exponents. you get 2.5 but generally people write it as an improper fraction

Answer 25

Oh okay. Thanks!

Answer 26

in other words, 2+ 1/2 is 4/2 + 1/2 = 5/2
and people write
\[ 8 a^\frac{5}{2} \]

Answer 27

Ya I got it. :)

Answer 28

if you are doing a lot of this, I would use the "fast way"
\[ (b^n)^m = b^{n \cdot m} \]
for example
\[ (4^1 \cdot a ^\frac{5}{3})^\frac{3}{2} \]
becomes
\[ 4^\frac{3}{2} \cdot a^{\frac{5}{3}\cdot \frac{3}{2} }\]

Answer 29

notice the expoinet of a is 5*3/(3*2) which simplifies to 5/2

to find what 4^3/2 is, we have to do either
\[ \left( 4^\frac{1}{2}\right)^3 = 2^3 = 8 \]
or
\[ \left( 4^3\right)^\frac{1}{2}= 64^\frac{1}{2} = 8 \]

Answer 30

Oh that makes sense.

Answer 31

if it doesn't , doing a dozen problems will be tedious.

Answer 32

Ya that's true.

Answer 33

Do you happen to know how to do this problem? How do you make both numbers have the same base? \[(5a ^{2/3})(4a ^{3/2})\]

Answer 34

you are multiplying 4 "things"
5* a^(2/3) * 4 * a^(3/2)
and you can switch the order:
5*4 * a^(2/3) * a^(3/2)

does that help to know how to tackle this ?

Answer 35

I think so. I got 20 as my final answer.

Answer 36

yes 4*5 is 20
but you can't ignore the a's (but you can multiply them using "same base, add exponents" rule)

Answer 37

Oh ya the a is \[a ^{1}\] So then the final answer would actually be 20a.

Answer 38

when you *multiply* (like here) , and you have the same base, you *add* the exponents.
in other words you get a^ new exponent

and the new exponent is 2/3 + 3/2

Answer 39

use a common denominator of 6 to get 4/6 + 9/6

Answer 40

Oh....I was multiplying them again. So that would equal 13/6.

Answer 41

yes. this stuff is confusing
but try to remember the x*x = x^2 or (putting in the exponents)
\[ x^1 \cdot x^1 = x^2\]
as an example of adding the exponents

Answer 42

Oh okay. So then it would equal \[20a ^{13/6}\]?

Answer 43

yes

Answer 44

Okay thanks again!