How do you simplify \frac { 2^ { 3} + 2^ { 7} + 2^ { 9} } { 2^ { 2} + 2^ { 6} + 2^ { 8} }?

2 Answers
Nov 3, 2017

2

Explanation:

Expresion = (2^3+2^7+2^9)/(2^2+2^6+2^8)

= (2^3(1+2^4+2^6))/(2^2(1+2^4+2^6))

= (2^3cancel((1+2^4+2^6)))/(2^2cancel((1+2^4+2^6)))

= 2^(3-2)= 2^1

=2

Nov 3, 2017

Factor out a 2 from the numerator to obtain an answer of 2.

Explanation:

Comparing the exponents in the numerator and denominator, we notice that the each of the exponents in the numerator are one greater than the exponents in the denominator (2^3=2^(1+2), 2^7=2^(1+6), 2^9=2^(1+8)).

We can use this pattern, along with the following law of exponents:

a^(x+y)=a^xa^y

to rewrite the fraction as (2*2^2+2*2^6+2*2^8)/(2^2+2^6+2^8).

From here, we can factor a 2 out in the numerator to obtain

2*(2^2+2^6+2^8)/(2^2+2^6+2^8)

and, since the numerator and denominator of the fraction on the right are equivalent, the fraction simply becomes 1, leaving us with 2*1=2 as our answer.