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Question Number 6877 by sarathon last updated on 01/Aug/16
$$\mathrm{the}\:\mathrm{bottom}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{of}\:\:{log}_{\frac{\mathrm{1}}{\mathrm{2}}} {x}\:\:\mathrm{wants}\:\mathrm{to}\:\mathrm{make}\:\:\mathrm{2} \\ $$$$\mathrm{2}\:\:\mathrm{how}?? \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 01/Aug/16
$${log}_{\frac{\mathrm{1}}{\mathrm{2}}} {x} \\ $$$$=\:{log}_{\mathrm{2}^{−\mathrm{1}} } {x} \\ $$$$=\:−\:\mathrm{1}\:{log}_{\mathrm{2}} ^{{x}} \\ $$$$=\:{log}_{\mathrm{2}} {x}^{−\mathrm{1}} \\ $$$$=\:{log}_{\mathrm{2}} ^{\frac{\mathrm{1}}{{x}}} \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 01/Aug/16
$${Let}\:\:\:{log}_{\frac{\mathrm{1}}{\mathrm{2}}} {x}={y} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{y}} ={x} \\ $$$$\left(\mathrm{2}^{-\mathrm{1}} \right)^{{y}} ={x} \\ $$$$\mathrm{2}^{-{y}} ={x} \\ $$$${log}_{\mathrm{2}} {x}=−{y} \\ $$$${y}=−{log}_{\mathrm{2}} {x}={log}_{\mathrm{2}} {x}^{-\mathrm{1}} \\ $$$${Hence}\:{log}_{\frac{\mathrm{1}}{\mathrm{2}}} {x}={log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$ \\ $$