How do you evaluate ${log}_{\frac{1}{7}} (\frac{1}{343})$ $log_(1/7) (1/343)$ ?

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Answer 1 · 2016-09-29T04:42:05

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Let $x = {log}_{\frac{1}{7}} (\frac{1}{343})$ $x=log_(1/7) (1/343)$

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

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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