anonymous
  • anonymous
How do you solve this problem? Please explain!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[(4a ^{5/3})^{3/2}\]
anonymous
  • anonymous
Solve for what?
anonymous
  • anonymous
It says to simplify in this problem

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IrishBoy123
  • 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
  • anonymous
How would I multiply the numbers inside the square root?
IrishBoy123
  • 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
  • anonymous
Okay. When you are multiplying exponents you have to add them right?
phi
  • phi
when multiplying the same base , you add the exponents.
anonymous
  • anonymous
So without the square root it equals 4a^5?
phi
  • 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
  • phi
yes, but you also have 4*4*4 to do
IrishBoy123
  • IrishBoy123
nah you also have 4 x 4 x 4!!
anonymous
  • anonymous
64a^5?
IrishBoy123
  • IrishBoy123
fine i'll leave it to @phi you are in great hands!! good luck
anonymous
  • anonymous
then it would equal 8a.....how would I square root the 5?
phi
  • 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
  • phi
\[ \sqrt{64 a^4 } \sqrt{a} \]
anonymous
  • anonymous
So the answer is \[8a \sqrt{a}\]?
phi
  • phi
8a^2 sqr(a)
anonymous
  • anonymous
So that would be simplified?
phi
  • phi
they probably want you to use exponents, so write sqr(a) as a^(1/2)
anonymous
  • anonymous
\[8a ^{2}a ^{1/2}\]
anonymous
  • anonymous
Like that?
phi
  • 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
  • anonymous
Oh okay. Thanks!
phi
  • phi
in other words, 2+ 1/2 is 4/2 + 1/2 = 5/2 and people write \[ 8 a^\frac{5}{2} \]
anonymous
  • anonymous
Ya I got it. :)
phi
  • 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
  • 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
  • anonymous
Oh that makes sense.
phi
  • phi
if it doesn't , doing a dozen problems will be tedious.
anonymous
  • anonymous
Ya that's true.
anonymous
  • 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
  • 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
  • anonymous
I think so. I got 20 as my final answer.
phi
  • 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
  • anonymous
Oh ya the a is \[a ^{1}\] So then the final answer would actually be 20a.
phi
  • 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
  • phi
use a common denominator of 6 to get 4/6 + 9/6
anonymous
  • anonymous
Oh....I was multiplying them again. So that would equal 13/6.
phi
  • 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
  • anonymous
Oh okay. So then it would equal \[20a ^{13/6}\]?
phi
  • phi
yes
anonymous
  • anonymous
Okay thanks again!

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