Difference between revisions of "2010 AMC 12B Problems/Problem 12"
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== Solution == | == Solution == | ||
<cmath> \log_{\sqrt{2}}\sqrt{x} + \log_2x + \log_4(x^2) + \log_8(x^3) + \log_{16}(x^4) = 40 </cmath> | <cmath> \log_{\sqrt{2}}\sqrt{x} + \log_2x + \log_4(x^2) + \log_8(x^3) + \log_{16}(x^4) = 40 </cmath> | ||
+ | \[\] | ||
<cmath> \frac{1}{2} \frac{\log_2x}{\log_2\sqrt{2}} + \log_2x + \frac{2\log_2x}{\log_24} + \frac{3\log_2x}{\log_28} + \frac{4\log_2x}{\log_216} = 40 </cmath> | <cmath> \frac{1}{2} \frac{\log_2x}{\log_2\sqrt{2}} + \log_2x + \frac{2\log_2x}{\log_24} + \frac{3\log_2x}{\log_28} + \frac{4\log_2x}{\log_216} = 40 </cmath> | ||
<cmath> \log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40 </cmath> | <cmath> \log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40 </cmath> | ||
<cmath> 5\log_2x = 40 </cmath> | <cmath> 5\log_2x = 40 </cmath> | ||
<cmath> \log_2x = 8 </cmath> | <cmath> \log_2x = 8 </cmath> | ||
− | <cmath> x = 256 (D) </cmath> | + | <cmath> x = 256 \;\; (D) </cmath> |
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} | {{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} |