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




Question Number 163060 by tounghoungko last updated on 03/Jan/22
  2log _3 ((x^2 /(27))) = 2+ ((log _3 ((1/x)))/(log _5 ((√x) )))
$$\:\:\mathrm{2log}\:_{\mathrm{3}} \left(\frac{{x}^{\mathrm{2}} }{\mathrm{27}}\right)\:=\:\mathrm{2}+\:\frac{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{log}\:_{\mathrm{5}} \left(\sqrt{{x}}\:\right)}\: \\ $$
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)))
$${log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)−{log}_{\mathrm{3}} \left(\mathrm{27}\right)=\mathrm{1}−\frac{{log}_{\mathrm{3}} {x}}{\mathrm{2}{log}_{\mathrm{5}} \sqrt{{x}}} \\ $$$${log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)−\mathrm{3}=\mathrm{1}−\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{5}} {x}} \\ $$$${log}_{\mathrm{5}} {x}=\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{3}} \mathrm{5}} \\ $$$$\Rightarrow{log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)−\mathrm{4}=−\frac{{log}_{\mathrm{3}} {x}}{\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{3}} \mathrm{5}}}=−{log}_{\mathrm{3}} \mathrm{5} \\ $$$$\Rightarrow{log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)=\mathrm{4}−{log}_{\mathrm{3}} \mathrm{5}= \\ $$$$={log}_{\mathrm{3}} \mathrm{3}^{\mathrm{4}} −{log}_{\mathrm{3}} \mathrm{5}={log}_{\mathrm{3}} \left(\frac{\mathrm{81}}{\mathrm{5}}\right) \\ $$$$\Rightarrow{x}=\sqrt{\frac{\mathrm{81}}{\mathrm{5}}}=\frac{\mathrm{9}}{\:\sqrt{\mathrm{5}}} \\ $$$${Ans}.\:{x}=\frac{\mathrm{9}}{\:\sqrt{\mathrm{5}}} \\ $$
Commented by mkam last updated on 03/Jan/22
how log_5 x = ((log_3 x)/(log_3 5)) ?
$${how}\:{log}_{\mathrm{5}} {x}\:=\:\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{3}} \mathrm{5}}\:? \\ $$
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))
$${b}={a}^{{log}_{{a}} {b}} =\left({c}^{{log}_{{c}} {a}} \right)^{{log}_{{a}} {b}} ={c}^{{log}_{{c}} {b}} \\ $$$$\Rightarrow{log}_{{c}} {b}={log}_{{c}} {a}\:×\:{log}_{{a}} {b} \\ $$$$\Rightarrow{log}_{{a}} {b}=\frac{{log}_{{c}} {b}}{{log}_{{c}} {a}} \\ $$

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