Difference between revisions of "1960 AHSME Problems/Problem 24"
Rockmanex3 (talk | contribs) (Solution for Problem 24) |
Rockmanex3 (talk | contribs) (Fix the format) |
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Rewrite the equation to <math>(2x)^x = 216</math>. Since <math>2x</math> is the base of the logarithm, we only need to check x-values that are positive. | Rewrite the equation to <math>(2x)^x = 216</math>. Since <math>2x</math> is the base of the logarithm, we only need to check x-values that are positive. | ||
− | With trial and error, <math>3</math> is a solution because <math>(2 \cdot 3)^3 = 6^3 = 216. Since < | + | With trial and error, <math>3</math> is a solution because <math>(2 \cdot 3)^3 = 6^3 = 216</math>. Since <math>(2x)^x</math> gets larger as x gets larger, <math>3</math> is the only solution, so the answer is <math>\boxed{\textbf{(A)}}</math>. |
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==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1960|num-b=23|num-a=25}} | {{AHSME 40p box|year=1960|num-b=23|num-a=25}} |
Latest revision as of 20:55, 11 May 2018
Problem
If , where is real, then is:
Solution
Rewrite the equation to . Since is the base of the logarithm, we only need to check x-values that are positive.
With trial and error, is a solution because . Since gets larger as x gets larger, is the only solution, so the answer is .
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AHSME Problems and Solutions |