Difference between revisions of "1968 AHSME Problems/Problem 23"
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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> | ||
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== Solution == | == Solution == | ||
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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> | ||
Line 22: | Line 17: | ||
<cmath>x^2+2x-3=x^2-2x-3</cmath> | <cmath>x^2+2x-3=x^2-2x-3</cmath> | ||
<cmath>x=0</cmath> | <cmath>x=0</cmath> | ||
− | However substituing into <math>\log(x-1)</math> gets a negative argument, which is impossible <math>\boxed{ | + | However substituing into <math>\log(x-1)</math> gets a negative argument, which is impossible <math>\boxed{B}</math>. |
~ Nafer | ~ Nafer |
Latest revision as of 21:21, 17 July 2024
Problem
If all the logarithms are real numbers, the equality
is satisfied for:
Solution
From the given we have
However substituing into
gets a negative argument, which is impossible
.
~ 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 |
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