How do you find the exact value of ${log}_{5} (\frac{1}{125})$ $log_5(1/125)$ ?

Question

11909 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-15T05:08:03

$- 3$ $-3$

${log}_{5} (\frac{1}{125}) = {log}_{5} (125^{- 1})$ $log_5(1/125) = log_5(125^-1)$

$= {log}_{5} {(5^{3})}^{- 1} = {log}_{5} 5^{- 3} = - 3 {log}_{5} 5 = - 3 (1) = - 3$ $= log_5 (5^3)^-1 = log_5 5^-3 = -3 log_5 5 = -3(1) = -3$

How do you find the exact value of log5(1125)log_5(1/125)?