- anonymous

How do you solve this problem? Please explain!

- jamiebookeater

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- anonymous

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

- anonymous

Solve for what?

- anonymous

It says to simplify in this problem

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## More answers

- IrishBoy123

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....

- anonymous

How would I multiply the numbers inside the square root?

- IrishBoy123

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.

- anonymous

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

- phi

when multiplying the same base , you add the exponents.

- anonymous

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

- phi

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)

- phi

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

- IrishBoy123

nah
you also have 4 x 4 x 4!!

- anonymous

64a^5?

- IrishBoy123

fine
i'll leave it to @phi
you are in great hands!! good luck

- anonymous

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

- phi

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

- phi

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

- anonymous

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

- phi

8a^2 sqr(a)

- anonymous

So that would be simplified?

- phi

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

- anonymous

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

- anonymous

Like that?

- phi

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

- anonymous

Oh okay. Thanks!

- phi

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

- anonymous

Ya I got it. :)

- phi

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
\[ 4^\frac{3}{2} \cdot a^{\frac{5}{3}\cdot \frac{3}{2} }\]

- phi

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
\[ \left( 4^3\right)^\frac{1}{2}= 64^\frac{1}{2} = 8 \]

- anonymous

Oh that makes sense.

- phi

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

- anonymous

Ya that's true.

- anonymous

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

- phi

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 ?

- anonymous

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

- phi

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

- anonymous

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

- phi

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

- phi

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

- anonymous

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

- phi

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

- anonymous

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

- phi

yes

- anonymous

Okay thanks again!

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