Find the exact value of logπ (1/ π5)

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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 

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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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