Difference between revisions of "1968 AHSME Problems/Problem 23"
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== Solution == | == Solution == | ||
− | <math>\fbox{}</math> | + | <math>\fbox{B}</math> |
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | From the given we have | ||
+ | <cmath>\log(x+3)+\log(x-1)=\log(x^2-2x-3)</cmath> | ||
+ | <cmath>\log(x^2+2x-3)=\log(x^2-2x-3)</cmath> | ||
+ | <cmath>x^2+2x-3=x^2-2x-3</cmath> | ||
+ | <cmath>x=0</cmath> | ||
+ | However substituing into <math>\log(x-1)</math> gets a negative argument, which is impossible <math>\boxed{D}</math>. | ||
+ | |||
+ | ~ Nafer | ||
== See also == | == See also == |
Revision as of 15:35, 24 December 2019
Contents
Problem
If all the logarithms are real numbers, the equality is satisfied for:
Solution
Solution 2
From the given we have However substituing into gets a negative argument, which is impossible .
~ Nafer
See also
1968 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 | ||
All AHSME Problems and Solutions |
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