Difference between revisions of "1969 AHSME Problems/Problem 17"
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== Solution == | == 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>. |
== See also == | == See also == |
Revision as of 16:28, 10 July 2015
Problem
The equation is satisfied by:
Solution
Let . Because
, the given expression can be rewritten as
. This can be factored as
, which has solutions
and
. Looking at the answer choices, we see that
is absent. Rewriting
as
and then applying the logarithm addition identity in reverse gives
. Applying the logarithm division identity shows that the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.