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Question Number 6373 by sanusihammed last updated on 25/Jun/16

If 9log_a ^5 = log_5 ^a find the value of a SOLUTION 9log_a ^5 = log_5 ^a Using the law of logarithm: log_b ^a = (1/(log_a ^b )) so, 9log_a ^5 = (1/(log_a ^5 )) Let log_a ^5 = h so, 9h = (1/h) cross multiply to get 9h^2 = 1 h^2 = 1/9 h = ± (√(1/9)) h = ± (1/3) h = (1/3) or h = −(1/3) Remember that log_a ^5 = h log_(a ) ^5 = (1/3) a^(1/3) = 5 Multiply both powers by 3 (a^(1/3) )^3 = 5^3 ∴ a = 125 Again log_a ^5 = −(1/3) a^(−(1/3)) = 5 multiply both powers by (−3) (a^(−(1/3)) )^(−3) = 5^(−3) ∴ a = (1/5^3 ) ∴ a = (1/(125)) Therefore, a = 125 or a = (1/(125)) DONE !

$${If}\:\:\:\:\mathrm{9}{log}_{{a}} ^{\mathrm{5}} \:\:=\:\:{log}_{\mathrm{5}} ^{{a}} \\ $$$${find}\:{the}\:{value}\:{of}\:{a}\: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{SOLUTION} \\ $$$$ \\ $$$$\mathrm{9}{log}_{{a}} ^{\mathrm{5}} \:\:=\:\:{log}_{\mathrm{5}} ^{{a}} \: \\ $$$$ \\ $$$${Using}\:{the}\:{law}\:{of}\:{logarithm}:\:\:{log}_{{b}} ^{{a}} \:=\:\frac{\mathrm{1}}{{log}_{{a}} ^{{b}} } \\ $$$${so},\:\:\mathrm{9}{log}_{{a}} ^{\mathrm{5}} \:=\:\frac{\mathrm{1}}{{log}_{{a}} ^{\mathrm{5}} } \\ $$$${Let}\:\:{log}_{{a}} ^{\mathrm{5}} \:=\:{h} \\ $$$${so},\:\mathrm{9}{h}\:=\:\frac{\mathrm{1}}{{h}} \\ $$$${cross}\:{multiply}\:{to}\:{get} \\ $$$$\mathrm{9}{h}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$${h}^{\mathrm{2}} \:=\:\mathrm{1}/\mathrm{9} \\ $$$${h}\:=\:\pm\:\sqrt{\frac{\mathrm{1}}{\mathrm{9}}} \\ $$$${h}\:=\:\pm\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${h}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:{h}\:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${Remember}\:{that} \\ $$$$ \\ $$$${log}_{{a}} ^{\mathrm{5}} \:=\:{h} \\ $$$${log}_{{a}\:} ^{\mathrm{5}} \:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} \:=\:\mathrm{5} \\ $$$${Multiply}\:{both}\:{powers}\:{by}\:\mathrm{3} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} \:=\:\mathrm{5}^{\mathrm{3}} \\ $$$$\: \\ $$$$\therefore\:\:\:\:{a}\:\:=\:\:\mathrm{125} \\ $$$$ \\ $$$${Again} \\ $$$${log}_{{a}} ^{\mathrm{5}} \:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}^{−\frac{\mathrm{1}}{\mathrm{3}}} \:=\:\:\mathrm{5} \\ $$$${multiply}\:{both}\:{powers}\:{by}\:\left(−\mathrm{3}\right) \\ $$$$\left({a}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right)^{−\mathrm{3}} =\:\mathrm{5}^{−\mathrm{3}} \\ $$$$ \\ $$$$\therefore\:\:\:\:\:{a}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} } \\ $$$$\therefore\:\:\:\:\:\:{a}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{125}} \\ $$$$ \\ $$$${Therefore}, \\ $$$$ \\ $$$${a}\:=\:\mathrm{125}\:\:{or}\:\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{125}} \\ $$$$ \\ $$$${DONE}\:! \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Answered by nburiburu last updated on 25/Jun/16

9 log_a 5 =log_5 a changing base 5 9((log_5 5)/(log_5 a))=log_5 a 9 =(log_5 a)^2 so log_5 a =+3 or log_5 a=−3 a= 5^3 or a=5^(−3)

$$\mathrm{9}\:{log}_{{a}} \mathrm{5}\:={log}_{\mathrm{5}} {a} \\ $$$${changing}\:{base}\:\mathrm{5} \\ $$$$\mathrm{9}\frac{{log}_{\mathrm{5}} \mathrm{5}}{{log}_{\mathrm{5}} {a}}={log}_{\mathrm{5}} {a} \\ $$$$\mathrm{9}\:=\left({log}_{\mathrm{5}} {a}\right)^{\mathrm{2}} \\ $$$${so} \\ $$$${log}_{\mathrm{5}} {a}\:=+\mathrm{3}\:\:\:\:{or}\:\:\:\:{log}_{\mathrm{5}} {a}=−\mathrm{3} \\ $$$${a}=\:\mathrm{5}^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:{or}\:\:\:\:\:\:{a}=\mathrm{5}^{−\mathrm{3}} \\ $$$$ \\ $$

Commented by sanusihammed last updated on 25/Jun/16

Thanks for your help

$${Thanks}\:{for}\:{your}\:{help} \\ $$

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