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Given that N > 1 and

\(\dfrac{1}{\log_2 N} + \dfrac{1}{\log_4 N} + \dfrac{1}{\log_6 N} + \dfrac{1}{\log_8 N} + \dfrac{1}{\log_{10} N} = \dfrac{1}{\log_x N}\)

find the value of the integer x.

 Dec 20, 2019
 #1
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Solve for x:
log(2)/log(N) + log(4)/log(N) + log(6)/log(N) + log(8)/log(N) + log(10)/log(N) = log(x)/log(N)

log(2)/log(N) + log(4)/log(N) + log(6)/log(N) + log(8)/log(N) + log(10)/log(N) = log(x)/log(N) is equivalent to log(x)/log(N) = log(2)/log(N) + log(4)/log(N) + log(6)/log(N) + log(8)/log(N) + log(10)/log(N):
log(x)/log(N) = log(2)/log(N) + log(4)/log(N) + log(6)/log(N) + log(8)/log(N) + log(10)/log(N)

Multiply both sides by log(N):
log(x) = log(2) + log(4) + log(6) + log(8) + log(10)

log(2) + log(4) + log(6) + log(8) + log(10) = log(2 4 6 8 10) = log(3840):
log(x) = log(3840)

Cancel logarithms by taking exp of both sides:

 

 x = 3840

 Dec 20, 2019

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