Difference between revisions of "2023 AIME I Problems/Problem 2"
Wuwang2002 (talk | contribs) m (→Solution 2 (extremely similar to above)) |
Wuwang2002 (talk | contribs) m (→Solution 2 (extremely similar to above)) |
||
Line 19: | Line 19: | ||
==Solution 2 (extremely similar to above)== | ==Solution 2 (extremely similar to above)== | ||
− | First, take the first equation and convert <math>\log_b\sqrt{n}</math> to <math>\log_b n^{\cfrac12}=\dfrac12\log_b n</math>. Square both sides to get <math>\log_b n=1/4 (\log_b n)^2</math>. Because a logarithm cannot be equal to <math>0</math>, <math>log_b n=4</math>. | + | First, take the first equation and convert <math>\log_b\sqrt{n}</math> to <math>\log_b n^{\cfrac12}=\dfrac12\log_b n</math>. Square both sides to get <math>\log_b n=1/4 (\log_b n)^2</math>. Because a logarithm cannot be equal to <math>0</math>, <math>\log_b n=4</math>. |
− | By another logarithm rule, <math>log_b(bn)=log_b b+log_b n=1+4=5</math>. Therefore, <math>4b=5</math>, and <math>b=\dfrac54</math>. Since <math>b^4=n</math>, we have <math>n=\dfrac{625}{256}</math>, and <math>a+b=\boxed{881}</math>. | + | By another logarithm rule, <math>\log_b(bn)=\log_b b+\log_b n=1+4=5</math>. Therefore, <math>4b=5</math>, and <math>b=\dfrac54</math>. Since <math>b^4=n</math>, we have <math>n=\dfrac{625}{256}</math>, and <math>a+b=\boxed{881}</math>. |
~wuwang2002 (feel free to remove if this is too similar to the above) | ~wuwang2002 (feel free to remove if this is too similar to the above) |
Revision as of 17:55, 8 February 2023
Problem
Positive real numbers and satisfy the equations The value of is where and are relatively prime positive integers. Find
Solution 1
Denote . Hence, the system of equations given in the problem can be rewritten as Thus, and . Therefore, Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (extremely similar to above)
First, take the first equation and convert to . Square both sides to get . Because a logarithm cannot be equal to , .
By another logarithm rule, . Therefore, , and . Since , we have , and .
~wuwang2002 (feel free to remove if this is too similar to the above)
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.