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2log-3-x-2-27-2-log-3-1-x-log-5-x-

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Question Number 163060 by tounghoungko last updated on 03/Jan/22
2log _3 ((x^2 /(27))) = 2+ ((log _3 ((1/x)))/(log _5 ((√x) )))
2log3(x227)=2+log3(1x)log5(x)
Commented by tounghoungko last updated on 04/Jan/22
Answered by aleks041103 last updated on 04/Jan/22
log_3 (x^2 )−log_3 (27)=1−((log_3 x)/(2log_5 (√x))) log_3 (x^2 )−3=1−((log_3 x)/(log_5 x)) log_5 x=((log_3 x)/(log_3 5)) ⇒log_3 (x^2 )−4=−((log_3 x)/((log_3 x)/(log_3 5)))=−log_3 5 ⇒log_3 (x^2 )=4−log_3 5= =log_3 3^4 −log_3 5=log_3 (((81)/5)) ⇒x=(√((81)/5))=(9/( (√5))) Ans. x=(9/( (√5)))
log3(x2)−log3(27)=1−log3x2log5xlog3(x2)−3=1−log3xlog5xlog5x=log3xlog35⇒log3(x2)−4=−log3xlog3xlog35=−log35⇒log3(x2)=4−log35==log334−log35=log3(815)⇒x=815=95Ans.x=95
Commented by mkam last updated on 03/Jan/22
how log_5 x = ((log_3 x)/(log_3 5)) ?
howlog5x=log3xlog35?
Commented by aleks041103 last updated on 03/Jan/22
b=a^(log_a b) =(c^(log_c a) )^(log_a b) =c^(log_c b) ⇒log_c b=log_c a × log_a b ⇒log_a b=((log_c b)/(log_c a))
b=alogab=(clogca)logab=clogcb⇒logcb=logca×logab⇒logab=logcblogca

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