1. ## simplify

1/ $3\sqrt{(x^3y^4)^2}$

2/ $\frac{(x^2y^4)^2}{x^5y^-1}$

3/ (x^3a+1)(x^a-3)/4sqrtx^8

2. Originally Posted by william

1/ $3\sqrt{(x^3y^4)^2}$

2/ $\frac{(x^2y^4)^2}{x^5y^-1}$

3/ $\frac{(x^{3a}+1)(x^a-3)}{4\sqrt{x^8}}$
Hello william,

[1] $3\sqrt{(x^3y^4)^2}=3x^3y^4$

[2] $\frac{(x^2y^4)^2}{x^5y^-1}=\frac{x^4y^8y}{x^5}=\frac{x^9}{x}$

[3] $\frac{(x^{3a}+1)(x^a-3)}{4\sqrt{x^8}}=\frac{x^{4a}-3x^{3a}+x^{a}-3}{4x^4}$

I made a few corrections on [3]. I'm not sure I translated your 3rd one correctly. Let me know.

3. Originally Posted by masters
Hello william,

[1] $\frac{(x^2y^4)^2}{x^5y^-1}=3x^3y^4$

[2] $\frac{(x^2y^4)^2}{x^5y^-1}=\frac{x^4y^8y}{x^5}=\frac{x^9}{x}$

[3] $\frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

I made a few corrections on [3]. I'm not sure I translated your 3rd one correctly. Let me know.
Thanks alot

What did you do there for 1, I am assuming you just wrote the second question as the first by mistake. For 3,

For 2 how did you get y^8 and not y^16, can you show steps please?

$\frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

in the numerator it is x^{3a+1} and x^{a-3}

and if you don't mind me asking how do i simplify (27)^-2/3

4. Originally Posted by william
Thanks alot

What did you do there for 1, I am assuming you just wrote the second question as the first by mistake. For 3,

$\frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

in the numerator it is x^{3a+1} and x^{a-3}

and if you don't mind me asking how do i simplify (27)^-2/3
William,

I don't have a clue what happened there. Some things did get out of whack, didn't they. Maybe you and I were editing at the same time and maybe divided by zero and clobbered the database.

[1] $3\sqrt{(x^3y^4)^2}=3x^3y^4$

5. Originally Posted by masters
William,

I don't have a clue what happened there. Some things did get out of whack, didn't they. Maybe you and I were editing at the same time and maybe divided by zero and clobbered the database.

[1] $3\sqrt{(x^3y^4)^2}=3x^3y^4$
I appreciate your replies but I am not sure how you are arriving at these answers

6. Originally Posted by william
I appreciate your replies but I am not sure how you are arriving at these answers
William,

Do you know that $\sqrt{a^2}=a$?

That's what I used to answer the first one. You could also look at it this way:

$\sqrt{a}=\sqrt{a^1}=a^{1/2}$. You can convert this into rational exponents by dividing the power in the radicand by the index of the root.

$3\sqrt{(x^3y^4)^2}=3(x^3y^4)^{2/2}=3(x^3 y^4)^1=3x^3 y^4$

Or, you could expand the radicand first and then simplify at the end like

$3 \sqrt{(x^3y^4)^2}=3\sqrt{x^6y^8}=3x^{6/2}y^{8/2}=3x^3y^4$

7. Originally Posted by masters
William,

Do you know that $\sqrt{a^2}=a$?

That's what I used to answer the first one. You could also look at it this way:

$\sqrt{a}=\sqrt{a^1}=a^{1/2}$. You can convert this into rational exponents by dividing the power in the radicand by the index of the root.

$3\sqrt{(x^3y^4)^2}=3(x^3y^4)^{2/2}=3(x^3 y^4)^1=3x^3 y^4$

Or, you could expand the radicand first and then simplify at the end like

$3 \sqrt{(x^3y^4)^2}=3\sqrt{x^6y^8}=3x^{6/2}y^{8/2}=3x^3y^4$
you have been great help

if i wanted to simplify (27)^{-2/3} would -8 be correct?

8. Originally Posted by william
Thanks alot

What did you do there for 1, I am assuming you just wrote the second question as the first by mistake. For 3,

For 2 how did you get y^8 and not y^16, can you show steps please?
Use this property of exponents: $(a^m)^n=a^{mn}$. So $(y^4)^2=y^{4 \cdot 2}=y^8$

Originally Posted by william

$\frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

in the numerator it is x^{3a+1} and x^{a-3}
Is it $\frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}$

If so, $\frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}=\frac{x^{(3a+3)+(a-3)}}{4x^4}=\frac{x^{4a}}{4x^4}=\frac{x^{4a-4}}{4}=\frac{x^{4(a-1)}}{4}$

Originally Posted by william
and if you don't mind me asking how do i simplify (27)^-2/3
$27^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{( \sqrt[3]{27})^2}=\frac{1}{(3)^2}=\frac{1}{9}$

9. Originally Posted by masters
Use this property of exponents: $(a^m)^n=a^{mn}$. So $(y^4)^2=y^{4 \cdot 2}=y^8$

Is it $\frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}$

If so, $\frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}=\frac{x^{(3a+3)+(a-3)}}{4x^4}=\frac{x^{4a}}{4x^4}=\frac{x^{4a-4}}{4}=\frac{x^{4(a-1)}}{4}$

$27^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{( \sqrt[3]{27})^2}=\frac{1}{(3)^2}=\frac{1}{9}$
Thank you very much for all the help you are a great mathematician and an excellent teacher. Cheers!