Difference between revisions of "1953 AHSME Problems/Problem 21"
(Created page with "==Problem== If <math>\log_{10} (x^2-3x+6)=1</math>, the value of <math>x</math> is: <math>\textbf{(A)}\ 10\text{ or }2 \qquad \textbf{(B)}\ 4\text{ or }-2 \qquad \textbf{(C)}...") |
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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=20|num-a=22}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:11, 27 November 2019
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 20 |
Followed by Problem 22 | |
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