Difference between revisions of "2023 IOQM/Problem 2"
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<cmath>\lbrace(a.b)\in N: 2 \leq a,b \leq2023,\:\: \log_{a}{b}+6\log_{b}{a}=5\rbrace</cmath> | <cmath>\lbrace(a.b)\in N: 2 \leq a,b \leq2023,\:\: \log_{a}{b}+6\log_{b}{a}=5\rbrace</cmath> | ||
− | ==Solution== | + | ==Solution quick== |
− | < | + | |
− | + | <math></math>assume \log_{a}{b}=x\$$ | |
+ | Then, | ||
+ | \begin{align*} | ||
+ | x + 5/x &= 6 & \text{because log} \\ | ||
+ | x^2 + 5 &= 6x & \text{by such-and-such} | ||
+ | \end{align*} | ||
+ | |||
+ | So, x equals to 2 or 3 | ||
+ | |||
+ | For 2, it's obvious that there are {2,4}...{44,1936} | ||
+ | |||
+ | For 3, it's obvious that there are {2,8}...{12,1728} | ||
+ | |||
+ | Thus, there are 43+11=53 pairs |
Revision as of 09:57, 2 October 2023
Problem
Find the number of elements in the set
Solution quick
$$ (Error compiling LaTeX. Unknown error_msg)assume \log_{a}{b}=x$$ Then, \begin{align*} x + 5/x &= 6 & \text{because log} \\ x^2 + 5 &= 6x & \text{by such-and-such} \end{align*}
So, x equals to 2 or 3
For 2, it's obvious that there are {2,4}...{44,1936}
For 3, it's obvious that there are {2,8}...{12,1728}
Thus, there are 43+11=53 pairs