Difference between revisions of "2005 AIME II Problems/Problem 5"

Latest revision as of 18:15, 4 April 2024

Problem

Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$

Solution 1

The equation can be rewritten as $\frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5$ Multiplying through by $\log a \log b$ and factoring yields $(\log b - 3\log a)(\log b - 2\log a)=0$ . Therefore, $\log b=3\log a$ or $\log b=2\log a$ , so either $b=a^3$ or $b=a^2$ .

For the case $b=a^2$ , note that $44^2=1936$ and $45^2=2025$ . Thus, all values of $a$ from $2$ to $44$ will work.
For the case $b=a^3$ , note that $12^3=1728$ while $13^3=2197$ . Therefore, for this case, all values of $a$ from $2$ to $12$ work.

There are $44-2+1=43$ possibilities for the square case and $12-2+1=11$ possibilities for the cube case. Thus, the answer is $43+11= \boxed{054}$ .

Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs $(a,b)$ , and not for the number of possible values of $b$ . Were the problem to ask for the number of possible values of $b$ , the values of $b^6$ under $2005$ would have to be subtracted, which would just be $2$ values: $2^6$ and $3^6$ . However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)

Solution 2

Let $k=\log_a b$ . Then our equation becomes $k+\frac{6}{k}=5$ . Multiplying through by $k$ and solving the quadratic gives us $k=2$ or $k=3$ . Hence $a^2=b$ or $a^3=b$ .

For the first case $a^2=b$ , $a$ can range from 2 to 44, a total of 43 values. For the second case $a^3=b$ , $a$ can range from 2 to 12, a total of 11 values.

Thus the total number of possible values is $43+11=\boxed{54}$ .

Solution 3(similar to solution 2)

Using the change of base formula on the second equation to change to base $a$ , we get $\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}$ . If we substitute $x$ for $\log_a(b)$ , we get $x + \frac{6}{x}$ . Multiplying by $x$ on both sides and solving, we get $x=3,2$ . Substituting back in, we get $\log_a(b) = 3,2$ . That means $a^3 = b$ or $a^2 = b$ . Since $b \leq 2005$ , we can see that for the cubed case, the maximum $a$ can be without exceeding 2005 is 12(because $13^3 = 2197$ ) and for the squared case it can be a maximum of 44. Since $a \neq 1$ , the number of values is $(44-1)+(12-1) = \boxed{54}$ .

~idk12345678

@@ Line 1: / Line 1: @@
 == Problem ==
+Determine the number of [[ordered pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>
-Determine the number of [[ordered pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>
+== Solution 1 ==
+The equation can be rewritten as <math> \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a  \log b}=5 </math> Multiplying through by <math>\log a \log b </math> and factoring yields <math>(\log b - 3\log a)(\log b - 2\log a)=0 </math>. Therefore, <math>\log b=3\log a </math> or <math>\log b=2\log a </math>, so either <math> b=a^3 </math> or <math> b=a^2 </math>.
+*For the case <math> b=a^2 </math>, note that <math> 44^2=1936 </math> and <math> 45^2=2025 </math>. Thus, all values of <math>a</math> from <math>2</math> to <math>44</math> will work.
+*For the case <math> b=a^3 </math>, note that <math> 12^3=1728 </math> while <math> 13^3=2197 </math>. Therefore, for this case, all values of <math>a</math> from <math>2</math> to <math>12</math> work.
+There are <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 43+11= \boxed{054}</math>.
+Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs <math>(a,b)</math>, and not for the number of possible values of <math>b</math>. Were the problem to ask for the number of possible values of <math>b</math>, the values of <math>b^6</math> under <math>2005</math> would have to be subtracted, which would just be <math>2</math> values:  <math>2^6</math> and <math>3^6</math>. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)
+==Solution 2 ==
+Let <math>k=\log_a b</math>. Then our equation becomes <math>k+\frac{6}{k}=5</math>. Multiplying through by <math>k</math> and solving the quadratic gives us <math>k=2</math> or <math>k=3</math>. Hence <math>a^2=b</math> or <math>a^3=b</math>.
+For the first case <math>a^2=b</math>, <math>a</math> can range from 2 to 44, a total of 43 values.
+For the second case <math>a^3=b</math>, <math>a</math> can range from 2 to 12, a total of 11 values.
+Thus the total number of possible values is <math>43+11=\boxed{54}</math>.
+==Solution 3(similar to solution 2)==
+Using the change of base formula on the second equation to change to base <math>a</math>, we get <math>\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}</math>. If we substitute <math>x</math> for <math>\log_a(b)</math>, we get <math>x + \frac{6}{x}</math>. Multiplying by <math>x</math> on both sides and solving, we get <math>x=3,2</math>. Substituting back in, we get <math>\log_a(b) = 3,2</math>. That means <math>a^3 = b</math> or <math>a^2 = b</math>. Since <math>b \leq 2005</math>, we can see that for the cubed case, the maximum <math>a</math> can be without exceeding 2005 is 12(because <math>13^3 = 2197</math>) and for the squared case it can be a maximum of 44. Since <math>a \neq 1</math>, the number of values is <math>(44-1)+(12-1) = \boxed{54}</math>.
-== Solution ==
+~idk12345678
-Note that the equation can be rewritten as: <math> \frac{\log b}{\log a}+6\frac{\log a}{\log b}=5 </math> or <math> \frac{(\log b)^2+6(\log a)^2}{\log a * \log b}=5 </math> Multiplying through by <math>\displaystyle \log a * \log b </math> and factoring yields <math> \displaystyle (\log b - 3\log a)(\log b - 2\log a)=0 </math>. Therefore, <math> \displaystyle \log b=3\log a </math> or <math>\displaystyle \log b=2\log a </math>. Therefore, either <math> b=a^3 </math> or <math> b=a^2 </math>. For the case <math> b=a^2 </math>, note that <math> 44^2=1936 </math> and <math> 45^2=2025 </math>. Thus, all values of <math>a</math> from 2 to 44 will work. For the case <math> b=a^3 </math>, note that <math> 12^3=1728 </math> while <math> 13^3=2197 </math>. Therefore, for this case, all values of <math>a</math> from 2 to 12 work. There are <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 43+11=054 </math>.
 == See also ==
-*[[Logarithm]]
 {{AIME box|year=2005|n=II|num-b=4|num-a=6}}
 [[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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