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Difference between revisions of "1967 AHSME Problems/Problem 4"

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== Solution ==
 
== Solution ==
−
<math>\fbox{C}</math>
+
We are given:
 +
<cmath>\frac{b^2}{ac} = x^y</cmath>
 +
 
 +
Taking the logarithm on both sides:
 +
<cmath>\log{\left(\frac{b^2}{ac}\right)} = \log{x^y}</cmath>
 +
 
 +
Using the properties of logarithms:
 +
<cmath>2\log{b} - \log{a} - \log{c} = y \log{x}</cmath>
 +
 
 +
Substituting the values given in the problem statement:
 +
<cmath>2q \log{x} - p \log{x} - r \log{x} = y \log{x}</cmath>
 +
 
 +
Since <math>x \neq 1</math>, dividing each side by <math>\log{x}</math> we get:
 +
<cmath>y = \boxed{\textbf{(C) } 2q - p - r}</cmath>
 +
 
 +
~ proloto
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1967|num-b=3|num-a=5}}   
+
{{AHSME 40p box|year=1967|num-b=3|num-a=5}}   
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:26, 28 September 2023

Problem

Given $\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}$, all logarithms to the same base and $x \not= 1$. If $\frac{b^2}{ac}=x^y$, then $y$ is:

$\text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr$


Solution

We are given: \[\frac{b^2}{ac} = x^y\]

Taking the logarithm on both sides: \[\log{\left(\frac{b^2}{ac}\right)} = \log{x^y}\]

Using the properties of logarithms: \[2\log{b} - \log{a} - \log{c} = y \log{x}\]

Substituting the values given in the problem statement: \[2q \log{x} - p \log{x} - r \log{x} = y \log{x}\]

Since $x \neq 1$, dividing each side by $\log{x}$ we get: \[y = \boxed{\textbf{(C) } 2q - p - r}\]

~ proloto

See also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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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