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1968 AHSME Problems/Problem 23: Difference between revisions

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


Line 10: Line 10:
\text{(D) no 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</math>
\text{(E) all real values of } x \text{ except } x=1</math>


== Solution ==
== Solution ==
<math>\fbox{B}</math>
== Solution 2 ==
From the given we have
From the given we have
<cmath>\log(x+3)+\log(x-1)=\log(x^2-2x-3)</cmath>
<cmath>\log(x+3)+\log(x-1)=\log(x^2-2x-3)</cmath>

Latest 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

From the given we have \[\log(x+3)+\log(x-1)=\log(x^2-2x-3)\] \[\log(x^2+2x-3)=\log(x^2-2x-3)\] \[x^2+2x-3=x^2-2x-3\] \[x=0\] However substituing into $\log(x-1)$ gets a negative argument, which is impossible $\boxed{B}$.

~ Nafer

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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