How do you evaluate ${log}_{\frac{2}{5}} (\frac{125}{8})$ $log_(2/5) (125/8)$ ?

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Answer 1 · 2016-11-24T02:18:06

$- 3$ $- 3$

Useful Hint for this question:

${log}_{a} a^{n} = n {log}_{a} a = n$ $log_aa^n=nlog_aa=n$

Solution :

${log}_{\frac{2}{5}} (\frac{125}{8}) = {log}_{\frac{2}{5}} {(\frac{5}{2})}^{3}$ $log_(2/5)(125/8)=log_(2/5)(5/2)^3$

= ${log}_{\frac{2}{5}} {(\frac{2}{5})}^{- 3}$ $log_(2/5)(2/5)^(-3)$

= $(- 3) {log}_{\frac{2}{5}} (\frac{2}{5})$ $(-3)log_(2/5)(2/5)$

= -3.

[As ${log}_{a} a = 1$ $log _a a=1$ ]

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