Instead of simplifying the whole terms, we can remove the terms which are common in both of the quantities and then compare the remaining parts

Let us remove the terms that are common on both sides
A: 1/8^2
B: (6*4)/(3^2 2^8 )
Lets rewrite A and B in terms of 2 and 3
A: 1/2^6
B: (2*3*2^2)/(3^2 2^8 )
Lets simplify the QTY B
A: 1/2^6
B: 2/3
Hence B is bigger.

let us reduce common terms on both sides:

LHS = 1/3^6 x 8^2
RHS = 6 x4/ 3^8 x 2^8

reduce again:
LHS = 1/8^2
RHS = 6 x4 / 9 x 2^8

LHS = 1/[2^3]^2
RHS = 6x 4/ 2^8

we get LHS denominator as = 1/2^6 and RHS = 6x 4/9 x 2^8
remove 1/2^6 on both sides
we get 1> 2/3
Hence A
Kudos if you like this post

Hey all, I reduced the numbers wherever I could to prime numbers so for example in Quantity A: 4^2 I made (2^2)^2 which is 2^4. Same thing with the 6^8 which in turn is 2^8 and 3^8. 5^3 is already in its prime form. Do the same thing to the 8^2 will convert it to 2^6. Using exponent rules take out 2^6 and 3^6 by subtracting from the numerator. What you're left with is 2^6 3^2 5^3= 72,000. Do the same thing for Quantity B and you should get 48,000. Thus my answer turned out to be that A was greater than B.

Seems to be differing opinions here. I tried to follow the posts but it was difficult for my head especially all the carrots and dashes. I think one of the ideas presented was to multiply essentially by 1, but in a different form, to make the two sides match. We can start by ignoring the 5 in each. Then multiply A by 6\frac{6[}{fraction], 4[fraction]4}, 9[fraction][/fraction]9 (3 squared over 3 squared), and 2 squared over 2 squared (for 2 to the sixth power in the denominator). I then could make them match except for 9 in the numerator (3 squared) and a 6 in the denominator, making the ratio 3:2 A:B

Upon completely simplifying by canceling out like terms in both quantities. You are left with
A = 1/(3^6 *2)
B = 1/(3^7) which is important to note = 1/ (3^6 * 3)

The denominator in B is larger than A which has the same numerator. Therefore, you know that A is the larger quantity.

Hope this helps.

cancel Q1 and Q2 by 6^8 *5^3 * 4^2 to obtain Q1. 1/(3^6 * 8^2) and Q2. 24/(3^8 * 2^8) OR 1/(3^7 * 2^5). Multiply both quantities by 3^6 to obtain Q1. 1/8^2=1/64 and Q2. 1/(3*32)=1/96. Q1>Q2 and answer is A.

Since there's disagreement I plugged it into a calculator. A = 72,000. B=48,000.

We can cancel \(5^3\) in both quantities and also eliminate \(6^8\) and \(4^2\) and \(3^6\) in Quantity A and B leaving behind

Quantity A

\(\dfrac{1}{8^2}\)

Quantity B

\(\dfrac{6^1 4^1}{3^2 2^8}\)



We can rewrite the denominator in both quantities as

Quantity A

\(2^3 2^3\)

Quantity B

\(3^2 2^3 2^3 2^2\)

We can then cancel \(2^3\) \(2^3\)on both sides, leaving behind

Quantity A

\(\dfrac{1}{1}\)

Quantity B

\(\dfrac{3 \times 2 \times 2 \times 2}{3 \times 3 \times 2 \times 2}\)

finally, we have

Quantity A

\(1\)

Quantity B

\(\dfrac{2}{3}\)

Thus Quantity A is greater.

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