2010 AMC 12B Problems/Problem 12

Revision as of 15:41, 15 February 2021 by Etvat (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

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 1

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

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

\[\log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40\]

\[5\log_2x = 40\]

\[\log_2x = 8\]

\[x = 256 \;\; (D)\]

Solution 2

Using the fact that $\log_{x^n}{y^n} = \log_{x}{y}$, we see that the equation becomes $\log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} = 40$. Thus, $5\log_{2}{x} = 40$ and $\log_{2}{x} = 8$, so $x = 2^8 = 256$, or $\boxed{(D)}$.


See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS