exponents- i don't understand them?

Question

how would u find the ratio of:

2^2001x 3^2003 /6^2002

Answer 1

i would break the denominator down from

6^2002 = 2^2002 * 3^2002

now you have

(2^2001 * 3^2003) / (2^2002 * 3^2002)

now that exponents from 2^ can subtract from each other and the same with 3, like this

2^(2001-2001) * 3^(2003-2002)
2^-1 * 3^1

= 3/2 = 1.5

hope that helps

Answer 2

6^2002 is the same as 2^2002 x 3^2002.
therefore if you split the bottom up like that
and you remember that when you divide the number, you subtract the indices
you'll have
2^2001 x 3^2003 divided by 2^2002 x 3^2002
which will give
2^-1 x 3^1
which is the same as a half times 3
which is one point five

Answer 3

In the numerator of the big fraction, we've got 2001 factors of 2 and 2003 factors of 3; while in the denominator we've got 2002 factors of six.

Since each factor of 6 in the denominator is a 2 and a 3, let's just remove 2001 factors of 6 in both places: that's 2001 2s and 2001 3s in the numerator, and 2001 6s in the denominator. That leaves us with:

3^2 / 6^1

Now remove one of the threes:

3 / 2

Answer 4

2^2001x 3^2003 /6^2002
= (2^2001)(3^2003)/(2*3)^2002
= (2^2001)(3^2003)/(2^2002)(3^2002)
= 3/2

We used the rules (xy)^m = x^my^m and
x^m/x^n = x ^(m-n)

Answer 5

This is "2" multiplied by itself 2001 times multiplied by 3 multiplied by itself 2003 times devided by 6 multiplied by itself 2002 times 2x3=6 take 2001 of each out 2 (2001 times) times 3 (2001 times) over 6 (2001 times) What you have left is 3x3 over6 (9/6) -=one and one half

Answer 6

(a*b)^c=(a^c)*(b^c)

(a^b)/(a^c)=a^(b-c)

2^2001x3^2003/6^2002 =
2^2001x3^2003/(2*3)^2002=
2^2001x3^2003/((2^2002)*3^2002)=

(2^2001/2^2002)*(3^2003/3^2002)=
1/2*3=3/2

Answer 7

2^2001 means 2*2*2*2*2..... 2001 times
to solve your expression, write it as:
(2^2001 * 3*3*3^2001)/6*6^2001
Now, 2*3=6, so, 2^2001*3^2001=6^2001
Cancel that out from the numerator and denominator and get:
(3*3)/6 = 9/6 = 1.5