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

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== Solution ==
 
== Solution ==
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<math>\fbox{D}</math>
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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_26</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>.
  
 
== See also ==
 
== See also ==

Latest revision as of 15:30, 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 $x=log_26$ 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 $\fbox{D}$.

See also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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