Exponents!

1. (3x^4)^2 = 6x^8      wrong as explained by     HighSchoolCalculus.

2. (3x^2y^4)(-2y^-8) = 9xy^4256y    

I will work this one through slowly for you then maybew you can present the answers again and someone will check them for you :)

\((3x^2y^4)(-2y^{-8})\\ =3x^2y^4\times-2y^{-8}\\ =3\times x^2\times y^4\times-2\times y^{-8}\\ =3\times-2\times x^2\times y^4\times y^{-8}\\ =-6\times x^2\times y^{4+-8}\\ =-6\times x^2\times y^{-4}\\ =\frac{-6\times x^2\times y^{-4}}{1}\\ =\frac{-6\times x^2}{y^{+4}}\\ =\frac{-6x^2}{y^{4}}\\ \)

3. -4/x^-6 = -4x^6    correct

4. (x^-2y^-4)^-3(y^4y^3) = x^6y^19    correct

\( (x^{-2}y^{-4})^{-3}(y^4y^3) = x^6y^{19}     \)

5. (x^2y^4)^4/(y^-2)^4 = x^6y^2

\((x^2y^4)^4/(y^{-2})^4 = x^6y^2\qquad wrong \)

? 6. (16x^2y^-4z^-1)^0/y^2 = 3/y^2 ?    wrong    -     Anything (except 0) raised to the power of 0 is 1

a = -2 b=6       huh?

7. (3ab^-1)^2 = 9a^2b       wrong

? 8.(4a^2b^-4)^0 ?

9.(x^5)^0=1      correct

?10.(a^2b^3)(missing exponent) = 1/a^2b^3 ?

11. x^-5*x^11=x^6       correct

12.w^8/w^5 = w^3       correct

Number one is actually 9x^8 (You multiplied 3 by 2 instead of squaring it, silly mistakes can be fatal)

By looking at most of the others you got most correct, though silly mistakes in some, try going through the step by step progress with them again.

Thanks for explaining everything Melody, I would have but my lazy bone acted up. Also Chemistry was calling my name.

Good day to you High School Calculus,

I thought that your aswer was quite good.  You gave some indication of what the problem was without giving full answers.

I encourage responses like this.  We give full answers far too often.

In this case I could really see from previous responses that Copead was really trying hard to learn from us so I decided to elaborate.

BUT

Like I said I thought your answer was good which why I gave you 5 stars.  :)

5. (x^2y^4)^4/(y^-2)^4 =

x^8 * y^16 / y^-8

But remember....a negative exponent "throws" the result back  across the fraction bar  and makes the exponent positive........so we have

x^8 * y^16 * y^8   =

x^8 * y^24

6. (16x^2y^-4z^-1)^0/y^2 =   1 / y^2      [assuming x, y and z are not = 0 ]

7. (3ab^-1)^2  =

3^2 * a^2 * b^-2  =

9a^2 / b^2

6. (16x^2y^-4z^-1)^0/y^2 = 1/y^2

8. Anything to the power of 0 is 1

Therefore equals to 1

10. \({{a}^{{2b}^{3}}}^{-1}= \frac{1}{{{a}^{{2b}^{3}}}}\)

Btw the question numbers i mentioned are Pythagorean triples xD