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Question Number 143812 by liberty last updated on 18/Jun/21

log _a (ax).log _x (ax)=log _a^2 ((1/a)) a>0 , a≠1 . So x = ?

$$\:\mathrm{log}\:_{\mathrm{a}} \left(\mathrm{ax}\right).\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{ax}\right)=\mathrm{log}\:_{\mathrm{a}^{\mathrm{2}} } \left(\frac{\mathrm{1}}{\mathrm{a}}\right) \\ $$ $$\:\mathrm{a}>\mathrm{0}\:,\:\mathrm{a}\neq\mathrm{1}\:.\:\mathrm{So}\:\mathrm{x}\:=\:? \\ $$

Answered by Olaf_Thorendsen last updated on 18/Jun/21

log_a (ax).log_x (ax) = log_a^2 ((1/a)) ((ln(ax))/(lna)).((ln(ax))/(lnx)) = −(1/2)log_a^2 a^2 = −(1/2) ln^2 (ax) = −(1/2)lna.lnx ln^2 a+ln^2 x+2lna.lnx = −(1/2)lna.lnx ln^2 x+ln^2 a+(5/2)lna.lnx = 0 (lnx+(5/4)lna)^2 = (9/(16))ln^2 a lnx+(5/4)lna= ±(3/4)∣lna∣ = ±(3/4)lna lnx= −(5/4)lna±(3/4)lna lnx= −2lna or −(1/2)lna x = (1/a^2 ) or (1/( (√a)))

$$\mathrm{log}_{{a}} \left({ax}\right).\mathrm{log}_{{x}} \left({ax}\right)\:=\:\mathrm{log}_{{a}^{\mathrm{2}} } \left(\frac{\mathrm{1}}{{a}}\right) \\ $$ $$\frac{\mathrm{ln}\left({ax}\right)}{\mathrm{ln}{a}}.\frac{\mathrm{ln}\left({ax}\right)}{\mathrm{ln}{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{{a}^{\mathrm{2}} } {a}^{\mathrm{2}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{ln}^{\mathrm{2}} \left({ax}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}{a}.\mathrm{ln}{x} \\ $$ $$\mathrm{ln}^{\mathrm{2}} {a}+\mathrm{ln}^{\mathrm{2}} {x}+\mathrm{2ln}{a}.\mathrm{ln}{x}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}{a}.\mathrm{ln}{x} \\ $$ $$\mathrm{ln}^{\mathrm{2}} {x}+\mathrm{ln}^{\mathrm{2}} {a}+\frac{\mathrm{5}}{\mathrm{2}}\mathrm{ln}{a}.\mathrm{ln}{x}\:=\:\mathrm{0} \\ $$ $$\left(\mathrm{ln}{x}+\frac{\mathrm{5}}{\mathrm{4}}\mathrm{ln}{a}\right)^{\mathrm{2}} =\:\frac{\mathrm{9}}{\mathrm{16}}\mathrm{ln}^{\mathrm{2}} {a} \\ $$ $$\mathrm{ln}{x}+\frac{\mathrm{5}}{\mathrm{4}}\mathrm{ln}{a}=\:\pm\frac{\mathrm{3}}{\mathrm{4}}\mid\mathrm{ln}{a}\mid\:=\:\pm\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}{a} \\ $$ $$\mathrm{ln}{x}=\:−\frac{\mathrm{5}}{\mathrm{4}}\mathrm{ln}{a}\pm\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}{a} \\ $$ $$\mathrm{ln}{x}=\:−\mathrm{2ln}{a}\:\mathrm{or}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}{a} \\ $$ $${x}\:=\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:\mathrm{or}\:\frac{\mathrm{1}}{\:\sqrt{{a}}} \\ $$

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