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sjuniv32
What is the sum of all possible value of x for which \(4^x – 2^{x + 3} – 2^{x + 2} + 32 = 0?\)

(A) 5

(B) 9

(C) 11

(D) 14

(E) 20

Solution:


    o \(4^x-2^{x+3}-2^{x+2} + 32 = 0\)
    o \(2^{2x} – 2^{x+3} – 2^{x+2} + 2^5 = 0\)
    o \(2^5 = 2^{x+3} + 2^{x+2} – 2^{2x}\)
    o \(2^5 = 2^{x}[2^3 + 2^2 – 2^x]\)
    o \(2^5 = 2^x[8+4-2^x]\)
    o \(2^5 = 2^x[12- 2^x]\)
    o \(2^5 = 2^x[3*2^2 – 2^x]\)
    o \(2^5 = 2^{x+2}[3 – 2^{x-2}]\)
Now, LHS is positive, the RHS will also be positive
    o We know that, \(2^{x+2}\) is always positive
      o \(3 – 2^{x-2}\) must be positive
      o \(2^{x-2} < 3\)
        o The only possible values of x are \(2\) and \(3\)
    o Thus, the required Sum is \(2 + 3 = 5\)
.
Hence, the correct answer is Option A.
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It’s important to use the many exponent rules both “backwards and forwards”

For each term, “back out” the multiplication of common bases and different exponents:


(4)^x = (2^2)^x = (2)^2x = (2)^x * (2)^x


- (2)^x + 3 = -(2)^x * (2)^3


- (2)^x + 2 = -(2)^x * (2)^2


And for the last term, we want to rearrange 32 to “match” the other terms we have rearranged:

32 = (2)^5 = (2)^2 * (2)^3


Putting all the rearranged terms together and setting them equal to 0, you can see the “hidden quadratic” (essentially the equations can be written as the product of 2 binomials) more clearly:


(2)^x * (2)^x - (2)^2 * (2)^x - (2)^3 * (2)^x + (2)^2 * (2)^3 = 0


(2^x - 2^2) * (2^x - 2^3) = 0


Rearranging the terms as the product of two factors that must equal zero, we can use the zero product rule/concept in that either one of the binomials must equal = 0

(2)^x = (2)^2

Or

(2)^x = (2)^3


Therefore

X = 2

Or

X = 3


Sum: 2 + 3 = 5

Answer (A)


Trickier question than the percentage correct suggests….

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