forget how to do this

## how do i simplify (2ab^-2c^-4)^-5 /(16a^-3bc^5)^2

# 5 Answers

(2ab^{-2}c^{-4})^{-5}/(16a^{-3}bc^{5})^{2} Given

(2^{-5}a^{-5}b^{10}c^{20})/(16^{2}a^{-6}b^{2}c^{10}) Distribute the exponents (multiply to find the power of a power)

(b^{10}c^{20}a^{6})/(16^{2}b^{2}c^{10*}2^{5}a^{5}) Negative exponents should be moved to the other part of the fraction

(a^{6}b^{10}c^{20})/(256*32a^{5}b^{2}c^{10}) Simplify powers of constants and arrange terms in alphabetic order

(a^{6}b^{10}c^{20})/(8192a^{5}b^{2}c^{10}) Multiply the constants

(a^{6}b^{10}c^{20}a^{-5}b^{-2}c^{-10})/8192 Move the lower exponent of each variable to the other part of the

fraction and change the sign

ab^{8}c^{10}/8192 Add exponents of like bases to simplify terms

(NOTE: You should see exponents (superscripts (and not ^)) throughout this explanation. If you are not seeing superscripts, the answer is not displaying correctly.)

Hi Mathalina.

PEMDAS says you should really start with P: inside the parenthesis first, but there's not much to do there. Then E: exponents.

In this problem you have an power raised to a power. There's a rule: (x^{m})^{n} = x^{mn}. All that means is when you have a power to a power, you MULTIPLY the exponents. It's like "distributing". But you have to be really careful
with the signs. And when you "distribute" that outer exponent, you must distribute to EVERYTHING inside the parenthesis, including the plain numbers (like the 2 in the numerator). Just be super careful with your signs.

So your equation becomes, after "distributing" the exponent (-5) on the top and the exponent (2) on the bottom:

( 2^{-5} * a^{-5 }* b^{10 }* c^{20 }) / (16^{2} * a^{-6} * b^{2} * c^{10} )

Now there are rules for exponents, but sometimes just simpler to see what is going all by making all the exponents positive. This is not the most direct way to solve this at this point, but it can be instructive. Remember that rule, x^{-n} = 1/x^{n},
which simply means you can flip something with an exponent from the top to the bottom (or vice versa) by changing the sign of the exponent. Let's do that now to the above equation by taking something with a negative exponent, flipping up to down, or down to
up, AND changing the sign of that exponent:

( a^{6} * b^{10} * c ^{20)} / (2^{5} * 16^{2}^{ }* a^{5} * b^{2} * c^{10
})

Let's just consider the a's. You have 6 lined up in a row all multiplied together on the top, and 5 in a row all multiplied together on the bottom. All but one will cancel, with the leftover one on the top:

(a*a*a*a*a*a) / (a*a*a*a*a) = a

Likewise with the b's: 10 all multiplied together on the top, 2 on the bottom, so after cancelling you have 8 left on the top.

By cancelling in this fashion you are left with:

(a^{1} * b^{8} * c^{10} ) / (2^{5} * 16^{2} )

Now use your calculator for the numbers (using the y^{x} or the ^ button (depending on your calculator), or just multiply them out) and you get:

(a b^{8} c ^{10} ) / (8192 )

Hope that helps.

Diane

If I understood it correctly, the expression is

[2a(b^-2)(c^-4)]^-5 / [16(a^-3)bc^5]^2

~~~~~~~~~

a^{m} • a^{n} = a^{m +}^{ n}

a^{m} ÷ a^{n} = a^{m - n}

(a^{m})^{n} = a^{m • n}

a^{- }^{n} = 1 / a^{n}

~~~~~~~~~~~~

(2ab^{-}^{ 2}c^{- 4})^{- 5}

(16a^{- 3}bc^{5})^{2}

2^{- 5} a^{- 5} b^{10} c^{20}

(2^{4})^{2} a^{- 6} b^{2} c^{10}

2^{- 5 - 8} a^{- 5}^{ - (- 6)} b^{10 - 2} c^{20 - 10}

2^{- 13} a^{1} b^{8} c^{10} = **(a b ^{8} c^{10}) / 2^{13}
**

# Comments

I love the approach of Nataliya D.

She always concise and precise.

As a fellow tutor, I highly recommend her.

Jonathan Yip

Cambridge, MA

Thank you, Mr. Yip. I try to pass on, what I learned from my math teachers: accuracy, briefness, pithiness and always the language of mathematics.

Any problem that involves simplifying exponents requires a good understanding of the exponent rules. There are several rules that you should know but I'll just go over the ones which we will use for this problem.

1. Product Rule: a^{n}a^{m} = a^{n+m}

2. Quotient Rule: a^{n}/a^{m }= a^{n-m}

3. Power Rule (#1): (ab)^{n} = a^{n}b^{n}

4. Power Rule (#2) (a^{n})^{m} = a^{nm}

5. Negative Exponent Rule: a^{-n} = 1/a^{n }or 1/a^{-n} = a^{n}. This is basically saying in order to make a negative exponent in the numerator positive, move that term to the denominator and vise versa.

With that said lets begin our simplification.

To start things off I would suggest combining rules 4 and 3 to the numerator and denominator separately. Doing so gives us:

(2^{-5}a^{-5}b^{10}c^{20})/(16^{2}a^{-6}b^{2}c^{10})

Next, using rule 5 we get:

(a^{6}b^{10}c^{20})/(16^{2}*2^{5}a^{5}b^{2}c^{10})

At this point we use rule 2 which gives us:

(ab^{8}c^{10})/(8192).

I hope this helps, let me know if you need me to elaborate more.

I would start this problem by taking care of the exponents on the outside of the parentheses.

Looking at just the numerator (top) of the fraction you have (2ab^-2c^-4)^-5. When you raise an exponent to an exponent, you multiply them. Keep in mind your rules for multiplying negatives. So you will get 2^-5a^-5b^10c^20.

Now look at the denominator (bottom) of the fraction. You have (16a^-3bc^5)^2. Again, when raising an exponent to another exponent, you multiply. You get 16^2a^-6b^2c^10.

So you now have 2^-5a^-5b^10c^20 / 16^2a^-6b^2c^10.

When you have a simple fraction that has negatives exponents you can easily fix the negative exponents by “changing their location”. If something in the numerator has a negative exponent with it, then move it to the denominator and drop the negative sign. Similarly, if something in the denominator has a negative exponent, then move it to the numerator and drop the negative sign. If anything has a positive exponent, then just leave it where it is.

So doing just that, you get a^6b^10c^20 / 2^5 16^2a^5b^2c^10. Now you have to reduce the fraction. Remember, when dividing, if you have the same variable on top and bottom you must subtract their exponents. So looking at just the a’s. You have a^6 on top and a^5 on bottom. So a^6/a^5 = a^(6-5) = a^1. Since the exponent is positive it stays on top. The easy trick for that is that whichever exponent is larger (top or bottom), the variable will stay there.

Doing the same for b’s, you get b^10 / b^2 = b^(10-2) = b^8, which stays on top. Next, c’s. c^20 / c^10 = c^(20-10) = c^10, which stays on top.

Lastly, looking at just the numbers you have 2^5 and 16^2 on bottom. 2^5 = 32 and 16^2 = 256. These are both on the bottom so you simply multiply them and you 32 x 256 = 8192.

So bringing all this together you have a^1b^8c^10 / 8192 .

## Comments

If you aren't sure about the rule for a power raised to a power when taking a test, you can go back to basics and figure it out or check your memory.

(a

^{3})^{2}= (a*a*a)^{2 }(a*a*a)^{2 }= (a*a*a) * (a*a*a)so...

(a

^{3})^{2}= a^{6}6 = 3*2

Just think about what those exponents mean and expand a simple example formula. (Be careful not to make it

toosimple. Don't use all 1's and 2's for your plug-in numbers because it might be a special case that wouldn't apply to all possible numbers. Use numbers that aren't multiples of each other.) With a little practice, you can do a lot of simple checks like this in your head. Knowing that you have the rule right takes a lot of the stress out of test taking!Gene

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