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Question Number 21082 by chernoaguero@gmail.com last updated on 12/Sep/17

Solve for x,   log_(0.2) (x+5) >0

$$\mathrm{Solve}\:\mathrm{for}\:{x},\:\:\:\mathrm{log}_{\mathrm{0}.\mathrm{2}} \left({x}+\mathrm{5}\right)\:>\mathrm{0} \\ $$

Answered by mrW1 last updated on 12/Sep/17

log _(0.2)  (x+5)=((ln (x+5))/(ln 0.2))>0  ln (x+5)<0  0<x+5<1  ⇒ −5<x<−4

$$\mathrm{log}\:_{\mathrm{0}.\mathrm{2}} \:\left(\mathrm{x}+\mathrm{5}\right)=\frac{\mathrm{ln}\:\left(\mathrm{x}+\mathrm{5}\right)}{\mathrm{ln}\:\mathrm{0}.\mathrm{2}}>\mathrm{0} \\ $$ $$\mathrm{ln}\:\left(\mathrm{x}+\mathrm{5}\right)<\mathrm{0} \\ $$ $$\mathrm{0}<\mathrm{x}+\mathrm{5}<\mathrm{1} \\ $$ $$\Rightarrow\:−\mathrm{5}<\mathrm{x}<−\mathrm{4} \\ $$

Commented bychernoaguero@gmail.com last updated on 12/Sep/17

Thank you

$${Thank}\:{you} \\ $$

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