Difference between revisions of "1969 AHSME Problems/Problem 17"

Revision as of 15:28, 10 July 2015

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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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == Solution ==
-<math>\fbox{D}</math>
+Let <math>2^x=a</math>. Because <math>2^(2x)=(2^x)^2</math>, the given expression can be rewritten as <math>a^2-8a+12=0</math>. This can be factored as <math>(a-6)(a-2)=0</math>, which has solutions <math>a=2^x=6</math> and <math>a=2^x=2</math>. Looking at the answer choices, we see that <math>x=1</math> is absent. Rewriting <math>2^x=6</math> as <math>x=log_2(6)</math> and then applying the logarithm addition identity in reverse gives <math>x=log_2(2)+log_2(3)=1+log_2(3)</math>. Applying the logarithm division identity shows that the answer is <math>\fbox{D}</math>.
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Difference between revisions of "1969 AHSME Problems/Problem 17"

Revision as of 15:28, 10 July 2015

Problem

Solution

See also