Difference between revisions of "1964 AHSME Problems/Problem 21"
Talkinaway (talk | contribs) (Created page with "== Problem 21== If <math>\log_{b^2}x+\log_{x^2}b=1, b>0, b \neq 1, x \neq 1</math>, then <math>x</math> equals: <math>\textbf{(A)}\ 1/b^2 \qquad \textbf{(B)}\ 1/b \qquad \te...") |
Talkinaway (talk | contribs) (→Solution) |
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You could inspect the equation here and see that <math>x=b</math> is one solution. Or, you can substitute <math>X = \ln x</math> and <math>B = \ln b</math> to get a quadratic in <math>X</math>: | You could inspect the equation here and see that <math>x=b</math> is one solution. Or, you can substitute <math>X = \ln x</math> and <math>B = \ln b</math> to get a quadratic in <math>X</math>: | ||
− | <math>X^2 | + | <math>X^2 + B^2 = 2BX</math> |
− | <math>X^2 - 2BX | + | <math>X^2 - 2BX + B^2 = 0</math> |
− | The above is a quadratic with coefficients <math>(1, -2B, | + | The above is a quadratic with coefficients <math>(1, -2B, B^2)</math>. Plug into the QF to get: |
<math>X = \frac{2B \pm \sqrt{4B^2 - 4B^2}}{2}</math> | <math>X = \frac{2B \pm \sqrt{4B^2 - 4B^2}}{2}</math> |
Revision as of 22:42, 23 July 2019
Problem 21
If , then equals:
Solution
Using natural log as a "neutral base", and applying the change of base formula to each term, we get:
You could inspect the equation here and see that is one solution. Or, you can substitute and to get a quadratic in :
The above is a quadratic with coefficients . Plug into the QF to get:
Either way, the answer is .
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.