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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 2)
 
(2 intermediate revisions by the same user not shown)
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\textbf{(E)}\ \sqrt{b} </math>     
 
\textbf{(E)}\ \sqrt{b} </math>     
  
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== Solution==
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== Solution 1==
  
 
Using natural log as a "neutral base", and applying the change of base formula to each term, we get:
 
Using natural log as a "neutral base", and applying the change of base formula to each term, we get:
Line 26: Line 26:
 
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>:
  
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<math>X^2 - B^2 = 2BX</math>
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<math>X^2 + B^2 = 2BX</math>
  
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<math>X^2 - 2BX - B^2 = 0</math>
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<math>X^2 - 2BX + B^2 = 0</math>
  
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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>
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Either way, the answer is <math>\boxed{\textbf{(D)}}</math>.
 
Either way, the answer is <math>\boxed{\textbf{(D)}}</math>.
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 +
==Solution 2==
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 +
All answers are of the form <math>x = b^n</math>, so we substitute that into the equation and try to solve for <math>n</math>.  We get:
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<math>\log_{b^2}x+\log_{x^2}b=1</math>
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<math>\log_{b^2}b^n + \log_{b^{2n}} b = 1</math>
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 +
By the definition of a logarithm, the first term on the left is asking for the exponent <math>x</math> needed to change the number <math>b^2</math> into <math>(b^2)^x</math> to get to <math>b^n</math>.  That exponent is <math>\frac{n}{2}</math>.
 +
 +
The second term is asking for a similar exponent needed to change <math>b^{2n}</math> into <math>b</math>.  That exponent is <math>\frac{1}{2n}</math>.
 +
 +
The equation becomes <math>\frac{n}{2} + \frac{1}{2n} = 1</math>.  Multiplying by <math>2n</math> gives the quadratic <math>n^2 + 1 = 2n</math>, which has the solution <math>n=1</math>.  Thus, <math>x = b^n = b^1</math>, and the answer is <math>\boxed{\textbf{(D)}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 01:45, 24 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 1

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 $\boxed{\textbf{(D)}}$.

Solution 2

All answers are of the form $x = b^n$, so we substitute that into the equation and try to solve for $n$. We get:

$\log_{b^2}x+\log_{x^2}b=1$

$\log_{b^2}b^n + \log_{b^{2n}} b = 1$

By the definition of a logarithm, the first term on the left is asking for the exponent $x$ needed to change the number $b^2$ into $(b^2)^x$ to get to $b^n$. That exponent is $\frac{n}{2}$.

The second term is asking for a similar exponent needed to change $b^{2n}$ into $b$. That exponent is $\frac{1}{2n}$.

The equation becomes $\frac{n}{2} + \frac{1}{2n} = 1$. Multiplying by $2n$ gives the quadratic $n^2 + 1 = 2n$, which has the solution $n=1$. Thus, $x = b^n = b^1$, and 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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