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How do you evaluate #log_(1/7) (1/343)#?

1 Answer

Sep 29, 2016

3

Explanation:

Let #x=log_(1/7) (1/343)#

#:. (1/7)^x =1/343#

Now consider: #1/7 =7^-1# and #1/343 = 1/7^3 = 7^-3#

Hence: #7^-x = 7^-3#

Equating exponents:

#-x =-3 -> x=3#

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