2010 AMC 12B Problems/Problem 12: Difference between revisions
Created page with '== Problem 12 == For what value of <math>x</math> does <cmath>\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?</cmath> <math>\textbf{(A)}\ 8 \…' |
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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> \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> | |||
\log_{\sqrt{2}}\sqrt{x} + \log_2x + \log_4(x^2) + \log_8(x^3) + \log_{16}(x^4) | <cmath> \log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40 </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} | <cmath> 5\log_2x = 40 </cmath> | ||
\log_2x + \log_2x + \log_2x + \log_2x + \log_2x | <cmath> \log_2x = 8 </cmath> | ||
5\log_2x | <cmath> x = 256 (D) </cmath> | ||
\log_2x | |||
x | |||
== 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}} | ||