Find the exact value of logπ (1/ π5)
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calendarEducator since 2008
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Given the logarithm expression:
log(n) (1/n^5)
We need to find the value.
First we will rewrite the quotient as a negative power
==> 1/n^5 = n^-5
==> log(n) (1/n^5) = log(n) n^-5
Now we know from logarithm properties that:
log a^b = b*log a
==> log(n) (1/n^5) = log(n) n^-5 = -5*log(n) n
Now we know that log(n) n = 1
==> log(n) 1/n^5 = -5*1 = -5
Then, the exact values of log(n) (1/n^5) = -5
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We need to find the exact value of log(n)[ 1/n^5], which is logarithm to the base n of 1/n^5.
Now log a^b = b* log a and this applies for logarithm to any base. Also, log (a) a = 1
log(n)[ 1/n^5]
we use 1/ n^5 = n^(-5)
=> log (n) [n^(-5)]
=> -5 * log (n) n
=> -5*1
=> -5
Therefore log(n)[ 1/n^5] = -5.
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