Difference between revisions of "2010 AMC 12B Problems/Problem 12"
(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}} |