Difference between revisions of "2010 AMC 12B Problems/Problem 12"
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| − | == Problem | + | == Problem == |
For what value of <math>x</math> does | For what value of <math>x</math> does | ||
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<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024</math> | <math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024</math> | ||
| − | == Solution == | + | == Solution 1== |
<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> | ||
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<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> | ||
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<cmath> 5\log_2x = 40 </cmath> | <cmath> 5\log_2x = 40 </cmath> | ||
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<cmath> \log_2x = 8 </cmath> | <cmath> \log_2x = 8 </cmath> | ||
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<cmath> x = 256 \;\; (D) </cmath> | <cmath> x = 256 \;\; (D) </cmath> | ||
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| + | ==Solution 2== | ||
| + | Using the fact that <math>\log_{x^n}{y^n} = \log_{x}{y}</math>, we see that the equation becomes <math>\log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} = 40</math>. Thus, <math>5\log_{2}{x} = 40</math> and <math>\log_{2}{x} = 8</math>, so <math>x = 2^8 = 256</math>, or <math>\boxed{(D)}</math>. | ||
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== See also == | == See also == | ||
| − | {{AMC12 box|year=2010|num-b= | + | {{AMC12 box|year=2010|num-b=11|num-a=13|ab=B}} |
| + | {{MAA Notice}} | ||
Latest revision as of 15:41, 15 February 2021
Contents
Problem
For what value of 
does
![\[\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?\]](http://latex.artofproblemsolving.com/e/0/c/e0c9ec94030e09b44a3086de323084adbecfb713.png)
Solution 1
![\[\log_{\sqrt{2}}\sqrt{x} + \log_2x + \log_4(x^2) + \log_8(x^3) + \log_{16}(x^4) = 40\]](http://latex.artofproblemsolving.com/7/4/9/7498c0fe81193b08592296dd3f814d2cbcdcf655.png)
![\[\frac{1}{2} \frac{\log_2x}{\log_2\sqrt{2}} + \log_2x + \frac{2\log_2x}{\log_24} + \frac{3\log_2x}{\log_28} + \frac{4\log_2x}{\log_216} = 40\]](http://latex.artofproblemsolving.com/f/7/f/f7fcc47e14a25755d1fdb4ab16fad12a4220b1f0.png)
![\[\log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40\]](http://latex.artofproblemsolving.com/2/4/b/24b4555144e83c662e5e0d838d6e4f66a883de4f.png)
![\[5\log_2x = 40\]](http://latex.artofproblemsolving.com/a/0/f/a0fa15d3c8038c8eeef4f27d4017db1935f5e66b.png)
![\[\log_2x = 8\]](http://latex.artofproblemsolving.com/e/6/9/e69d0058e5864841add8ab316c7bd9daf5355459.png)
![\[x = 256 \;\; (D)\]](http://latex.artofproblemsolving.com/0/f/1/0f19d42c2e2ac72ae9ab1b20718de822d0c5a168.png)
Solution 2
Using the fact that 
, we see that the equation becomes 
. Thus, 
and 
, so 
, or 
.
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| 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. 
