Need simplification for the maths problem

## How to solve this maths problem (4a^-2 b)^2 (3a^3 b^-4)^-2

# 2 Answers

Since this is all multiplication and division, (no addition or subtraction),

you can simplify like variable parts.

You will need to use the fact that a^{m} = a^{m-n} ( for a not = 0)

a^{n}

AND, you need to know about negative exponents:

1 = a^{n} , a^{-n} =
b^{m}

a^{-n } b^{-m} a^{n}

Or, think about any thing with a negative exponent can be moved to the other side of the fraction as that base to a positive exponent. (numerator to denominator, denominator to numerator)

Also, if you have (a^{m}b^{n})^{k} = a^{mk}b^{nk} (Or, you want to mulitply exponents here)

Now that we have some of the rules, there are a number of different ways

you could start to simplify this problem, but here is one way:

What if we apply the 2 and -2 exponents to what's inside the parantheses, respectively.

(4a^{-2}b)^{2} is the same as 4^{2}a^{-4}b^{2} (Remember, you can think of 4 as 4^{1}, and you can multiply exponents.

(3a^{3}b^{-4})^{-2} is the same as what? (again, multiply the exponents)

After multiplying exponents, then you can eliminate any negative exponents by

putting those terms in the denominator with positive exponents.

Can you finish it from here?

(4a^{-2} b)^{2} (3a^{3} b^{-4})^{-2} = 4^{2} a^{-4} b^{2} 3^{-2} a^{-6} b^{8} = (16/9) a^{-10} b^{10} = (16/9)(b/a)^{10}

## Comments

AND, when you have the same base and are multiplying, you will add exponents:

For example, x

^{3}x^{4}= x^{7}(in general, a^{m}a^{n}= a^{m+n})You will need to use this rule when you combine with one of the variables in this problem

Ok, did you find the answer yet?

(3a

^{3}b^{-4})^{-2}= 3^{-2}a^{-6}b^{8}So, you have (4

^{2}a^{-4}b^{2})(3^{-2}a^{-6}b^{8}). Since, you have removed the outside exponents, this is just all multiplication, which we can re-order as:4

^{2}•3^{-2}•a^{-4}•a^{-6}•b^{2•}b^{8}4

^{2}3^{-2}a^{-10}b^{10}So, now let's get rid of negative exponents:

4

^{2}b^{10}= 16b^{10}3

^{2}a^{10}9a^{10}Thank you very much for the response Mr George.highly recommend you as a teacher !!!

Comment