Maths help?

Question

Can anyone work out:
(2^n * 8^n) / (2^2n * 16)

in the form of 2^an+b

hope that makes sense
Thanks for the help

Answer 1

Write everything as a power of 2:

8^n = 2^3n

16 = 2^4

So, the equation becomes (2^n * 2^3n)/(2^2n * 2^4)

Simplify the numerator and denominator by adding exponents: 2^4n / 2^(2n + 4)

Subtract exponents for the division: 2^(4n - 2n - 4)

You are left with 2^(2n - 4)

Answer 2

(2^n * 8^n) / (2^2n * 16) = 1/4
or
2^n * 2³n over 2²n * 2^4
Simplifying:
um the exponents of equal bases:
n + 3n over 2n + 4
2^4n over2^2n+4
2² over 2^4 = 2²-4 = 2-² = 1/4
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Answer 3

= (2^n x (2³)^n) / (2^2n x 2^4)

= (2^n x 2^3n) / 2^(2n + 4)

= 2^(4n) / 2^(2n + 4)

= 2^(2n - 4)

This is in the form 2^(an + b) where a = 2 and b = - 4

Answer 4

I can help you work this out.

One general rule about exponents is that the exponents add if you have the same base.

5^a * 5^b = 5^(a+b)

another general rule about exponents in fractions is that the denominator is equivalent to having a negative exponent.

1/5 = 5^(-1), and 1/(5^2) = 5^(-2)

One final rule you'll need is to realize that 8 and 16 are also powers of two. The rule you'll need is this:
(5^2)^3 = 5^(2*3)

Now try to use these rules creatively to solve your problem.