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Difference between revisions of "1964 AHSME Problems/Problem 21"

(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...")
 
(→‎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 - B^2 = 2BX</math>
+
<math>X^2 + B^2 = 2BX</math>
  
−
<math>X^2 - 2BX - B^2 = 0</math>
+
<math>X^2 - 2BX + B^2 = 0</math>
  
−
The above is a quadratic with coefficients <math>(1, -2B, -B^2)</math>.  Plug into the QF to get:
+
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 21:42, 23 July 2019

Problem 21

If $\log_{b^2}x+\log_{x^2}b=1, b>0, b \neq 1, x \neq 1$, then $x$ equals:

$\textbf{(A)}\ 1/b^2 \qquad \textbf{(B)}\ 1/b \qquad \textbf{(C)}\ b^2 \qquad \textbf{(D)}\ b \qquad \textbf{(E)}\ \sqrt{b}$

Solution

Using natural log as a "neutral base", and applying the change of base formula to each term, we get:

$\frac{\ln x}{\ln b^2} + \frac{\ln b}{\ln x^2} = 1$


$\frac{\ln x}{2\ln b} + \frac{\ln b}{2\ln x} = 1$


$\frac{\ln x \ln x + \ln b \ln b}{2\ln b \ln x} = 1$


$\ln x \ln x + \ln b \ln b = 2\ln b \ln x$

You could inspect the equation here and see that $x=b$ is one solution. Or, you can substitute $X = \ln x$ and $B = \ln b$ to get a quadratic in $X$:

$X^2 + B^2 = 2BX$

$X^2 - 2BX + B^2 = 0$

The above is a quadratic with coefficients $(1, -2B, B^2)$. Plug into the QF to get:

$X = \frac{2B \pm \sqrt{4B^2 - 4B^2}}{2}$

$X = B$

$\ln x = \ln b$

$x = b$

Either way, the answer is $\boxed{\textbf{(D)}}$.

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

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