Alg 2 Question...?

Question

I have absolutely no idea how to do this... could you plz explain... THANKS!

How do I Simplify:

(a^4/5 * c^-1/2)^4

Write your answer without using negative exponents.
Assume that all variables are positive real numbers.

Answer 1

first you have to open the bracket.
for this, multiply each exponent with 4

so u get it as a^16/5 * c^-2

now, we are asked to get an answer without negative exponent. Here, c is having a negative exponent .. so get it to the denominator. so, -2 will become +2. the final answer will be

a^16/5
----------
c^2

note that the a term is in numerator and c term is in denominator.

Answer 2

I'm going to ellaborate in detail, for fun and so you can see exactly whats going on and why.

There are many concepts going on in your problem. For example:

One concept is "a" and "c". These are simply "variables" or numbers that you do not know. By TRADITIONALLY letters a, b and c are unknown numbers that you are not trying to solve and therefore don't care what they equal. This is as opposed to x and y, for example. Those you would typically solve for, and care what they are. You might even hear a, b and c referred to as "constants".

The next concept is parentheses "()". Do what's in the parentheses first. It's not really relevant in this problem, but it is good to be aware of.

Next is multiplication "*". IMPORTANT CONCEPT! ;) You need to know that raising a multiplication to a power is different than raising an addition to a power. In other words:
(a * c)^2
[Pronounced "the quantity a times c, all to the power of 2" or "a times c squared"]
is different than (a + c)^2.
[Pronounced "the quantity a plus c, all to the power of 2" or "a plus c squared"]
They are different in the way that you solve them, not in the grand scheme of how powers work.
[cuz, of course, how powers work is, a*a means a^2 and a*a*a means a^3.]
In that example, the simplification for (a * c)^2 is a^2 * c^2.

This leads us into the concern of your problem.
"a" is also known as "a^1".
[Pronounced "a to the first"]

This means that a^2 is the same as (a^1)^2.
This is not the schoolroom explanation and proof of why this works, it's just my way of illustrating. Your teacher prolly explained it differently.

SO... while a^2 is a^2, what is (a^1)^2?
If given (a^1)^2, we were to think hmmm should I multiply the 1 with the 2? or should I raise 1 to the power of 2?
Let's think. If we raise 1 to the power of 2, that means we are multiplying 1 times 1. The result would be a^1. Smack me with a trout if a^2 = a^1. Oops, I mean, since they are not equal, then the other option, to multiply the inside power with the outside power, is the case.

Sigh. Having said all that, we now can see that to solve a power to a power, you multiply the inside power with the outside power.

Lets look at your example information, but break it up a little bit. Lets look at just (a^4/5)^4 (I'll write that 4 below as 4 over 1 so you can picture multiplying across).

(4/5) * (4/1) = (4 * 4)/(5 * 1) = 16/5

(a^4/5)^4 therefore equals a^16/5.

Go take a break and come back.
Welcome back. Did you beat that pesky mirror match round of World of Warcraft?

What do we know so far?
We know how to raise one factor that is to a power to another power. But how do we raise two factors to a power? First we will quickly raise the "c" factor to the power of 4. Then we will look at how to multiply the two new factors together.

Next is the part with the c. Dont worry yet what the minus sign means. Just treat it like a number to multiply with 4. So...

To solve (c^-1/2)^4, again we multiply (-1/2) with 4.

(-1/2) * (4/1) = (-1 * 4)/(2 * 1) = -4/2 = -2

Therefore (c^-1/2)^4 = c^-2.

This next part is explained on math.com, but it is not named, and I forgot what the concept is called. It is the part where we see that we are allowed to raise powers of multiplication factors within the parentheses by a power outside the parentheses without having to do anything fancy.

(xy)^a = x^a * y^a

Push the expando button to see this in terms of our given problem...
(x^1 * y^1)^a = (x^1)^a * (y^1)^a = x^a * y^a

In this case, the x above refers to your a^4/5, the y above refers to c^-1/2, and the a above refers to your 4 outside of the parentheses.

Lemme see if I can expand this next part, so you dont get confused. This is the actual solving of the problem so listen up, Beavis.

original problem:
(a^4/5 * c^-1/2)^4

by the math.com postulate:
(a^4/5 * c^-1/2)^4 = (a^4/5)^4 * (c^-1/2)^4
= a^16/5 * c^-2

Now we tackle that bit given in the question:
"Write your answer without using negative exponents."

On the same math.com I gave you before, you will find that:
x^(-a) = 1 / x^a

Right. So...
our c^-2 that we found above can be written as 1/c^2

By multiplication:
a^16/5 * c^-2 is also known as:
(a^16/5) * 1/(c^2) is also known as:

(a^16/5)/(c^2)

As my teacher would say, "Circle your answer... Full credit."



The part of your problem that says, "Assume that all variables are positive real numbers." is given so the problem will work for technical reasons that are not necessary to understand to solve this problem. In other words, you dont need to write anything on your answer sheet to satisfy that section of the problem. Or as my friend Danny Rizzo (RIP) would say, "Just say 'OK!' "

Answer 3

= a^(4/5)^4 x c^(-2)
= a^(16/5) x 1 / c²

Answer 4

(a^(16/5))/(c^2)