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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-

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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 !
If9loga5=log5afindthevalueofaSOLUTION9loga5=log5aUsingthelawoflogarithm:logba=1logabso,9loga5=1loga5Letloga5=hso,9h=1hcrossmultiplytoget9h2=1h2=1/9h=±19h=±13h=13orh=−13Rememberthatloga5=hloga5=13a13=5Multiplybothpowersby3(a13)3=53∴a=125Againloga5=−13a−13=5multiplybothpowersby(−3)(a−13)−3=5−3∴a=153∴a=1125Therefore,a=125ora=1125DONE!
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)
9loga5=log5achangingbase59log55log5a=log5a9=(log5a)2solog5a=+3orlog5a=−3a=53ora=5−3
Commented by sanusihammed last updated on 25/Jun/16
Thanks for your help
Thanksforyourhelp

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