Difference between revisions of "1968 AHSME Problems/Problem 23"
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== Solution 2 == | == 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 14:35, 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
![\[\log(x+3)+\log(x-1)=\log(x^2-2x-3)\]](http://latex.artofproblemsolving.com/1/f/5/1f506a774fc52fe7ea88131c01fb3c5f3515901a.png)
![\[\log(x^2+2x-3)=\log(x^2-2x-3)\]](http://latex.artofproblemsolving.com/1/e/e/1eedaa202b5117fd0d2ca10fa853ae762ca5bf43.png)
![\[x^2+2x-3=x^2-2x-3\]](http://latex.artofproblemsolving.com/6/c/d/6cd45e9ecb70c334f2eafbee324601876c896bd9.png)
![\[x=0\]](http://latex.artofproblemsolving.com/0/e/5/0e5065ea42e71241d1571fd868bae3267c53d818.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
