Difference between revisions of "1968 AHSME Problems/Problem 23"

Revision as of 20:21, 17 July 2024

Problem

If all the logarithms are real numbers, the equality $\log(x+3)+\log(x-1)=\log(x^2-2x-3)$ is satisfied for:

$\text{(A) all real values of }x \quad\\ \text{(B) no real values of } x\quad\\ \text{(C) all real values of } x \text{ except } x=0\quad\\ \text{(D) no real values of } x \text{ except } x=0\quad\\ \text{(E) all real values of } x \text{ except } x=1$

Solution

$\fbox{B}$

1968 AHSC (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 • 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.

@@ Line 13: / Line 13: @@
 == Solution ==
 <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{B}</math>.
-~ Nafer
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Difference between revisions of "1968 AHSME Problems/Problem 23"

Revision as of 20:21, 17 July 2024

Problem

Solution

See also