Question 202091
Simplify these following expressions:
(2x)^4 x^3
(-3y)^4
(3z)^2(6z^2)^-3
5x2⁄25x5
10(x+y)^4/5(x+y)^3
1/6x(3x^2)^3
(5x^2)^3(1/25x^4)^2
(2z^2)^-5*z^-10
(3/x)^4(4/x)^-2
[2(r-s)]^2/(r-s)^3




Let's take a few at a time......


I see your first problem as this:


{{{(2x)^4}}}*{{{ x^3}}}

If that is the problem, then you multiply:

2 to the 4th power -- 2*2*2*2 = 16  (You distribute the exponent "4" to the 2 and to the "x"....



and then you multiply {{{x^4}}}*{{{x^3}}}.  Since the bases are the same, you ADD the exponents, so you have {{{x^7}}}


FINAL answer:   {{{16x^7}}}



Next:(-3y)^4


That is:  -3*-3*-3*-3 = 81  (See how the exponent 4 is distributed to the -3 and to the y?)
{{{y^4}}} is the second part of the answer


FINAL answer:  {{{81y^4}}}



Next:  (3z)^2(6z^2)^-3



If the problem is this:  {{{(3z)^2}}}*{{{6z^2}}} raised to the power of -3
That is {{{9z^2}}} times {{{6^-3}}}times {{{z^-6}}}


That becomes:   {{{9z^2}}} *  {{{1/6^3z^6}}}


Why?  The exponent "2" distributes to the 3 and to the z.


The exponent -3 distributes to the 6 and to the "z".  HOWEVER, for the "z", because you are raising a power to a power, you have to MULTIPLY the exponents. Therefore, the "z" becomes {{{z^-6}}} or.... {{{1/z^6}}}


The problem is now:


{{{9z^2}}}/{{{216z^6}}}


THAT becomes


{{{1/24 z^4}}}  (we reduced the fraction {{{9/216}}} to {{{1/24}}}.





Let's stop here for a second and go over some exponent rules, k?   Then if you need help with your other answers, ask again.  It seems like you may not understand exponents...........



{{{x^2}}}* {{{x^3}}} = {{{x^5}}}  When multiplying... if the bases are the same, ADD the exponents


{{{x^3}}}/{{{x^2}}} = {{{x}}}   When dividing... if the bases are the same, SUBTRACT the exponents


If you raise a power to a power, then MULTIPLY the exponents.  


If you have a NEGATIVE exponent in the NUMERATOR, just put it in the denominator and make it a POSITIVE exponent:


{{{x^-3}}} is really:   {{{1/x^3}}}


If you have a NEGATIVE EXPONENT in the DENOMINATOR, just put it in the numerator and make it a POSITIVE exponent.

{{{1/x^-3}}} is really:  {{{x^3}}}



I hope this helps......