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Difference between revisions of "2003 AIME I Problems/Problem 4"

m (→‎Problem: You wrote log(n)-10 when it was log(n)-1)
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Given that <math> \log_{10} \sin x + \log_{10} \cos x = -1 </math> and that <math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1), </math> find <math> n. </math>
 
Given that <math> \log_{10} \sin x + \log_{10} \cos x = -1 </math> and that <math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1), </math> find <math> n. </math>
  
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== Solution ==
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== Solution 1 ==
 
Using the properties of [[logarithm]]s, we can simplify the first equation to <math>\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1 </math>. Therefore, <cmath> \sin x \cos x = \frac{1}{10}.\qquad (*)</cmath>
 
Using the properties of [[logarithm]]s, we can simplify the first equation to <math>\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1 </math>. Therefore, <cmath> \sin x \cos x = \frac{1}{10}.\qquad (*)</cmath>
  
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By the Pythagorean identities, <math>\sin ^2 x + \cos ^2 x = 1</math>, and we can substitute the value for <math>\sin x \cos x</math> from <math>(*)</math>. <math>1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012} </math>.
 
By the Pythagorean identities, <math>\sin ^2 x + \cos ^2 x = 1</math>, and we can substitute the value for <math>\sin x \cos x</math> from <math>(*)</math>. <math>1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012} </math>.
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 +
== Solution 2 ==
 +
Examining the first equation, we simplify as the following:
 +
<cmath>\log_{10} \sin x \cos x = -1</cmath>
 +
<cmath>\implies \sin x \cos x = \frac{1}{10}</cmath>
 +
 +
With this in mind, examining the second equation, we may simplify as the following (utilizing logarithm properties):
 +
<cmath>\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - \log_{10} 10)</cmath>
 +
<cmath>\implies \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} \frac{n}{10})</cmath>
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<cmath>\implies \log_{10} (\sin x + \cos x) = \log_{10} \sqrt{\frac{n}{10}}</cmath>
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 +
From here, we may divide both sides by <math>\log_{10} (\sin x + \cos x)</math> and then proceed with the change-of-base logarithm property:
 +
<cmath>1 = \frac{\log_{10} \sqrt{\frac{n}{10}}}{\log_{10} (\sin x + \cos x)}</cmath>
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<cmath>\implies 1 = \log_{\sin x + \cos x} \sqrt{\frac{n}{10}}</cmath>
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 +
Thus, exponentiating both sides results in <math>\sin x + \cos x = \sqrt{\frac{n}{10}}</math>. Squaring both sides gives us
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<cmath>\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{n}{10}</cmath>
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 +
Via the Pythagorean Identity, <math>\sin^2 x + \cos^2 x = 1</math> and <math>2\sin x \cos x</math> is simply <math>\frac{1}{5}</math>, via substitution. Thus, substituting these results into the current equation:
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<cmath>1 + \frac{1}{5} = \frac{n}{10}</cmath>
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<cmath>\implies \frac{6}{5} = \frac{n}{10}</cmath>
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Using simple cross-multiplication techniques, we have <math>5n = 60</math>, and thus <math>\boxed{n = 012}</math>.
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~ nikenissan
  
 
== See also ==
 
== See also ==

Revision as of 10:27, 13 July 2020

Problem

Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$

Solution 1

Using the properties of logarithms, we can simplify the first equation to $\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1$. Therefore, \[\sin x \cos x = \frac{1}{10}.\qquad (*)\]

Now, manipulate the second equation. \begin{align*} \log_{10} (\sin x + \cos x) &= \frac{1}{2}(\log_{10} n - \log_{10} 10) \\ \log_{10} (\sin x + \cos x) &= \left(\log_{10} \sqrt{\frac{n}{10}}\right) \\ \sin x + \cos x &= \sqrt{\frac{n}{10}} \\ (\sin x + \cos x)^{2} &= \left(\sqrt{\frac{n}{10}}\right)^2 \\ \sin^2 x + \cos^2 x +2 \sin x \cos x &= \frac{n}{10} \\ \end{align*}

By the Pythagorean identities, $\sin ^2 x + \cos ^2 x = 1$, and we can substitute the value for $\sin x \cos x$ from $(*)$. $1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012}$.

Solution 2

Examining the first equation, we simplify as the following: \[\log_{10} \sin x \cos x = -1\] \[\implies \sin x \cos x = \frac{1}{10}\]

With this in mind, examining the second equation, we may simplify as the following (utilizing logarithm properties): \[\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - \log_{10} 10)\] \[\implies \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} \frac{n}{10})\] \[\implies \log_{10} (\sin x + \cos x) = \log_{10} \sqrt{\frac{n}{10}}\]

From here, we may divide both sides by $\log_{10} (\sin x + \cos x)$ and then proceed with the change-of-base logarithm property: \[1 = \frac{\log_{10} \sqrt{\frac{n}{10}}}{\log_{10} (\sin x + \cos x)}\] \[\implies 1 = \log_{\sin x + \cos x} \sqrt{\frac{n}{10}}\]

Thus, exponentiating both sides results in $\sin x + \cos x = \sqrt{\frac{n}{10}}$. Squaring both sides gives us \[\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{n}{10}\]

Via the Pythagorean Identity, $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x$ is simply $\frac{1}{5}$, via substitution. Thus, substituting these results into the current equation: \[1 + \frac{1}{5} = \frac{n}{10}\] \[\implies \frac{6}{5} = \frac{n}{10}\]

Using simple cross-multiplication techniques, we have $5n = 60$, and thus $\boxed{n = 012}$. ~ nikenissan

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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