Difference between revisions of "1977 AHSME Problems/Problem 18"

(Created page with "== Problem 18 == If <math>y=(\log_23)(\log_34)\cdots(\log_n[n+1])\cdots(\log_{31}32)</math> then <math>\textbf{(A) }4<y<5\qquad \textbf{(B) }y=5\qquad \textbf{(C) }5<y<6\qq...")
 
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Note that <math>\log_ab = \dfrac{\logb}{\loga}</math>. Then <math>y=(\dfrac{\log3}{\log2})(\dfrac{\log4}{\log3})\cdots(\dfrac{\log32}{\log31}) = \dfrac{\log32}{\log2} = \log_232 = \boxed{\text{(B) }y=5}</math>.
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Note that <math>\log_{a}b = \dfrac{\log{b}}{\log{a}}</math>. Then <math>y=(\dfrac{\log3}{\log2})(\dfrac{\log4}{\log3})\cdots(\dfrac{\log32}{\log31}) = \dfrac{\log32}{\log2} = \log_232 = \boxed{\text{(B) }y=5}</math>.

Latest revision as of 12:05, 22 November 2016

Problem 18

If $y=(\log_23)(\log_34)\cdots(\log_n[n+1])\cdots(\log_{31}32)$ then

$\textbf{(A) }4<y<5\qquad \textbf{(B) }y=5\qquad \textbf{(C) }5<y<6\qquad \textbf{(D) }y=6\qquad \\ \textbf{(E) }6<y<7$


Solution

Solution by e_power_pi_times_i


Note that $\log_{a}b = \dfrac{\log{b}}{\log{a}}$. Then $y=(\dfrac{\log3}{\log2})(\dfrac{\log4}{\log3})\cdots(\dfrac{\log32}{\log31}) = \dfrac{\log32}{\log2} = \log_232 = \boxed{\text{(B) }y=5}$.