2010 AMC 12B Problems/Problem 12: Difference between revisions

Revision as of 12:54, 31 May 2011

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 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

Revision as of 17:53, 9 July 2010 view source Lg5293 (talk \| contribs) 64 edits No edit summary ← Older edit		Revision as of 12:54, 31 May 2011 view source Btilm305 (talk \| contribs) Administrators 292 edits →See also: fixed box numbers Newer edit →
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	== See also ==		== See also ==
	{{AMC12 box\|year=2010\|num-b=12\|num-a=14\|ab=B}}		{{AMC12 box\|year=2010\|num-b=11\|num-a=13\|ab=B}}