# calculate log 5 (1/125) ( 5 is under the log)

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log 5 (1/125)

We know that 125 = 5^3

Substitute :

==> log 5 (1/125) = log 5 (1/5^3)

Now we know that 1/5^3 = 5^-3

==> log 5 (1/125) = log 5 (5^-3)

= -3*log 5 (5)

= -3 * 1= -3

The answer is (-3)

log5 (1/125)

Solution:

Since 5 is under log , we pressume log 5 is as good as logarithm of 5 to the base 10.

Then log 5 (1/125) = log5+ log (1/125) = log5 + log (1/5^3), log(ab) = log a+logb , log (a/b) = log a- logb , logarithm properties.

= log5+log(1) - log5^3 =

= log5 +0 -3log5, as log(1) = 0 and log a^b = bloga.

=log5 -3log 5

= -2log5

= - 2*0.690970004

= -1.397940009.

If the 5 is base of logarithm, then log5 (1/125) = log5 (1/5^3) = log5 (5^(-3) ) = -3 , as logk (k^m) = m, by definition of logarithm.

Assuming that 5 is the base of the logarithm, we'll calculate the logarithm, using the quotient property:

log 5 (1/125) = log 5 (1) - log 5 (125)

We could write the following terms in this way:

log 5 (1) = log 5 (5^0)

We'll use the power property of log:

log 5 (1) = log 5 (5^0) = 0* log 5 (5) = 0

log 5 (125) = log 5 (5^3) = 3* log 5 (5) = 3*1 = 3

** log 5 (1/125) = log 5 (1) - log 5 (125) = 0 - 3 = -3**