Menu Close

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-




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

Leave a Reply

Your email address will not be published. Required fields are marked *