I was wondering how that would be simplified. I can only have positive exponents in the end. Can you please walk me through it?

## (a ^2 *b^2 *c ^ -3) *^ -4 *a^4 *c^-3 and that whole thing is divided by 2a^-2 *b^3

# 3 Answers

From the fact that for m and n,integers, and a, real number, (a^{m})^{n} = a^{mn}, (a^{m}b^{n})^{p} = a^{mp}b^{np,}

a^{m} * a^{n} = a^{m+n}, a^{m} / a^{n} = a^{m-n},

(a^{2}b^{2}c^{-3})^{-4} * a^{4}c^{-3 }= a^{-8}b^{-8}c^{12}a^{4}c^{-3} = a^{-4}b^{-8}c^{9}

So, ^{ }[(a^{2}b^{2}c^{-3})^{-4 }* a^{4}c^{-3}]/2a^{-2}b^{3} = a^{-4}b^{-8}c^{9}/2a^{-2}b^{3} = a^{-2}b^{-11}c^{9}/2 =
c^{9}/2a^{2}b^{11}

I'm going to write this in a little differently because it is how I am interpreting your question, hopefully I am reading the same thing you trying to get answered.

(a^{2}·b^{2}·c^{-3}) ^{-4}· a^{4}·c^{-3} ÷ (2a^{-2}· b^{3})

We need to know rules for exponents. Combine like exponents (the exponents for x's together, etc.)

an exponent raised to an exponent: multiply the exponents (x^{2})^{4} = x^{8}

an exponent multiplied to and exponent: add the exponents x^{2}·x^{4} = x^{6}

an exponent divided by an exponent: sub the bottom from the top x^{2}/x^{4} = x^{-2}

a positive exponent means it belongs on the top of a fraction

a negative exponent means it belongs on the bottom of a fraction

Lets break it into two pieces and put back together at end

#1: (a^{2}·b^{2}·c^{-3}) ^{-4}· a^{4}·c^{-3}

distribute the -4 to the ( )

a^{-8} b^{-8} c^{12} · a^{4}c^{-3}

add exponents (because all the letters are being multiplied together)

a^{-4} b^{-8} c^{9}

put the exponents on top and bottom of a fraction based on + or - sign or exponent

c^{9}/a^{4}b^{8}

we will come back to this in a second

#2

2a^{-2}· b^{3}

put the exponents on top and bottom of a fraction based on + or - sign or exponent

2b^{3}/a^{2}

#1 and #2 together

(c^{ 9}/a^{4}b^{ 8})÷(2b^{ 3}/a^{2})

to divide fractions we flip the bottom one (the reciprocal) and multiply the fractions together

(c ^{9}/a^{4}b ^{8})×(a^{ 2}/2b
^{3})

multiply across the top to get new numerator, and across bottom to get new denominator

a^{2}c ^{9}/2a^{4}b^{8}b^{ 3}

add, sub, multiply as needed to simplify further

a: 2-4 = -2

b: 8+3 = 11

c: 9-0 = 9

c ^{9}/2a^{2}b^{11}

(a ^2 *b^2 *c ^ -3) *^ -4 *a^4 *c^-3 divided by 2a^-2 *b^3

I'm not sure what the *^ is supposed to mean. I'm going to assume that your problem looks like this:

[(a^{2}b^{2}c^{-3})^{-4}a^{4}c^{-3}]/2a^{-2}b^{3}

The only operations are exponents, multiplication, and division.

This means we can bring the denominator up to the numerator if we change the sign of the exponents:

(a^{2}b^{2}c^{-3})^{-4}a^{4}c^{-3}*2a^{2}b^{-3}

We want to get rid of the parentheses to an exponent. Remember (A^{m}B^{n})^{w} = A^{mw}B^{nw}

a^{-8}b^{-8}c^{12}a^{4}c^{-3}2a^{2}b^{-3}

Like bases, add exponents:

2a^{(-8+4+2)}b^{(-8-3)}c^{(12-3)}

2a^{-2}b^{-11}c^{9}

That could be the answer, but you said you only want positive exponents. So, now we just put the terms with the negative exponents in the denominator and change their sign.

2c^{9}/a^{2}b^{11}

Good luck on your test tomorrow.

## Comments

the two should be in the denominator here.

Good catch. I will correct my answer.

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