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}}

Revision as of 17:52, 9 July 2010

Problem 12

For what value of $x$ does

\[\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?\]

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024$

Solution

\[\log_{\sqrt{2}}\sqrt{x} + \log_2x + \log_4(x^2) + \log_8(x^3) + \log_{16}(x^4)  = 40\] \[\] \[\frac{1}{2} \frac{\log_2x}{\log_2\sqrt{2}} + \log_2x + \frac{2\log_2x}{\log_24} + \frac{3\log_2x}{\log_28} + \frac{4\log_2x}{\log_216}  = 40\] \[\log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40\] \[5\log_2x = 40\] \[\log_2x = 8\] \[x = 256 \;\; (D)\]

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 12 Problems and Solutions