Question

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

Answer 1

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`

Answer 2

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.