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

Answer 1

Abigail Allen

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#

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Answer 2

Ethan Adkins

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

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.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.