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

Revision as of 17:53, 9 July 2010

For what value of $x$ does

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

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

$\log_{\sqrt{2}}\sqrt{x} + \log_2x + \log_4(x^2) + \log_8(x^3) + \log_{16}(x^4) = 40$

$\frac{1}{2} \frac{\log_2x}{\log_2\sqrt{2}} + \log_2x + \frac{2\log_2x}{\log_24} + \frac{3\log_2x}{\log_28} + \frac{4\log_2x}{\log_216} = 40$

$\log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40$

$5\log_2x = 40$

$\log_2x = 8$

$x = 256 \;\; (D)$

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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 == 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> \log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40 </cmath>
 <cmath> 5\log_2x = 40 </cmath>
 <cmath> \log_2x = 8 </cmath>
 <cmath> x = 256 \;\; (D) </cmath>
 == See also ==
 {{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}