Difference between revisions of "1953 AHSME Problems/Problem 21"
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==Solution== | ==Solution== | ||
We know that <math>x^2-3x+6=10^1</math>, after expanding the logarithm. Taking the 10 to the other side, <math>x^2-3x-4=0</math>. Factoring, we get <math>(x-4)(x+1)=0</math>, so <math>x=\boxed{4,-1\Rightarrow \text{(D)}}</math>. | We know that <math>x^2-3x+6=10^1</math>, after expanding the logarithm. Taking the 10 to the other side, <math>x^2-3x-4=0</math>. Factoring, we get <math>(x-4)(x+1)=0</math>, so <math>x=\boxed{4,-1\Rightarrow \text{(D)}}</math>. | ||
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+ | ==See Also== | ||
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+ | {{AHSME 50p box|year=1953|num-b=18|num-a=20}} | ||
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+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 21:27, 15 January 2018
Problem
If , the value of is:
Solution
We know that , after expanding the logarithm. Taking the 10 to the other side, . Factoring, we get , so .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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