1969 AHSME Problems/Problem 17

Problem

The equation $2^{2x}-8\cdot 2^x+12=0$ is satisfied by:

$\text{(A) } log(3)\quad \text{(B) } \tfrac{1}{2}log(6)\quad \text{(C) } 1+log(\tfrac{3}{2})\quad \text{(D) } 1+\frac{log(3)}{log(2)}\quad \text{(E) none of these}$

Solution

Let $2^x=a$ . Because $2(2x)=(2^x)^2$ , the given expression can be rewritten as $a^2-8a+12=0$ . This can be factored as $(a-6)(a-2)=0$ , which has solutions $a=2^x=6$ and $a=2^x=2$ . Looking at the answer choices, we see that $x=1$ is absent. Rewriting $2^x=6$ as $x=log_2(6)$ and then applying the logarithm addition identity in reverse gives $x=log_2(2)+log_2(3)=1+log_2(3)$ . Applying the logarithm division identity shows that the answer is $\fbox{D}$ .

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1969 AHSME Problems/Problem 17

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