Question Number 6373 by sanusihammed last updated on 25/Jun/16
$${If}\:\:\:\:\mathrm{9}{log}_{{a}} ^{\mathrm{5}} \:\:=\:\:{log}_{\mathrm{5}} ^{{a}} \\ $$$${find}\:{the}\:{value}\:{of}\:{a}\: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{SOLUTION} \\ $$$$ \\ $$$$\mathrm{9}{log}_{{a}} ^{\mathrm{5}} \:\:=\:\:{log}_{\mathrm{5}} ^{{a}} \: \\ $$$$ \\ $$$${Using}\:{the}\:{law}\:{of}\:{logarithm}:\:\:{log}_{{b}} ^{{a}} \:=\:\frac{\mathrm{1}}{{log}_{{a}} ^{{b}} } \\ $$$${so},\:\:\mathrm{9}{log}_{{a}} ^{\mathrm{5}} \:=\:\frac{\mathrm{1}}{{log}_{{a}} ^{\mathrm{5}} } \\ $$$${Let}\:\:{log}_{{a}} ^{\mathrm{5}} \:=\:{h} \\ $$$${so},\:\mathrm{9}{h}\:=\:\frac{\mathrm{1}}{{h}} \\ $$$${cross}\:{multiply}\:{to}\:{get} \\ $$$$\mathrm{9}{h}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$${h}^{\mathrm{2}} \:=\:\mathrm{1}/\mathrm{9} \\ $$$${h}\:=\:\pm\:\sqrt{\frac{\mathrm{1}}{\mathrm{9}}} \\ $$$${h}\:=\:\pm\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${h}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:{h}\:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${Remember}\:{that} \\ $$$$ \\ $$$${log}_{{a}} ^{\mathrm{5}} \:=\:{h} \\ $$$${log}_{{a}\:} ^{\mathrm{5}} \:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} \:=\:\mathrm{5} \\ $$$${Multiply}\:{both}\:{powers}\:{by}\:\mathrm{3} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} \:=\:\mathrm{5}^{\mathrm{3}} \\ $$$$\: \\ $$$$\therefore\:\:\:\:{a}\:\:=\:\:\mathrm{125} \\ $$$$ \\ $$$${Again} \\ $$$${log}_{{a}} ^{\mathrm{5}} \:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}^{−\frac{\mathrm{1}}{\mathrm{3}}} \:=\:\:\mathrm{5} \\ $$$${multiply}\:{both}\:{powers}\:{by}\:\left(−\mathrm{3}\right) \\ $$$$\left({a}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right)^{−\mathrm{3}} =\:\mathrm{5}^{−\mathrm{3}} \\ $$$$ \\ $$$$\therefore\:\:\:\:\:{a}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} } \\ $$$$\therefore\:\:\:\:\:\:{a}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{125}} \\ $$$$ \\ $$$${Therefore}, \\ $$$$ \\ $$$${a}\:=\:\mathrm{125}\:\:{or}\:\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{125}} \\ $$$$ \\ $$$${DONE}\:! \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by nburiburu last updated on 25/Jun/16
$$\mathrm{9}\:{log}_{{a}} \mathrm{5}\:={log}_{\mathrm{5}} {a} \\ $$$${changing}\:{base}\:\mathrm{5} \\ $$$$\mathrm{9}\frac{{log}_{\mathrm{5}} \mathrm{5}}{{log}_{\mathrm{5}} {a}}={log}_{\mathrm{5}} {a} \\ $$$$\mathrm{9}\:=\left({log}_{\mathrm{5}} {a}\right)^{\mathrm{2}} \\ $$$${so} \\ $$$${log}_{\mathrm{5}} {a}\:=+\mathrm{3}\:\:\:\:{or}\:\:\:\:{log}_{\mathrm{5}} {a}=−\mathrm{3} \\ $$$${a}=\:\mathrm{5}^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:{or}\:\:\:\:\:\:{a}=\mathrm{5}^{−\mathrm{3}} \\ $$$$ \\ $$
Commented by sanusihammed last updated on 25/Jun/16
$${Thanks}\:{for}\:{your}\:{help} \\ $$