Question Number 80844 by jagoll last updated on 07/Feb/20
$${If}\:\left(\mathrm{log}_{\mathrm{10}} \left({ax}\right)\right)\left(\mathrm{log}_{\mathrm{10}} \left({bx}\right)\right)=−\mathrm{1} \\ $$$${find}\:{x}\:{in}\:{term}\:{a}\:{and}\:{b}\: \\ $$
Commented by jagoll last updated on 07/Feb/20
$${can}\:{anyone}\:{help}\:{me} \\ $$
Commented by jagoll last updated on 07/Feb/20
$$\frac{{b}}{{a}}\geqslant…\: \\ $$$${it}\:{the}\:{answer} \\ $$
Answered by mr W last updated on 07/Feb/20
$${solution}\:{exists}\:{only}\:{for}\:{a},{b}\:>\mathrm{0}\:{or}\:{a},{b}<\mathrm{0}. \\ $$$${assume}\:{a},{b}>\mathrm{0}\:{in}\:{following}. \\ $$$$\mathrm{log}\:\left({ax}\right)\:×\:\mathrm{log}\:\left({bx}\right)=−\mathrm{1} \\ $$$$\left(\mathrm{log}\:{a}+\mathrm{log}\:{x}\right)\left(\mathrm{log}\:{b}+\mathrm{log}\:{x}\right)=−\mathrm{1} \\ $$$$\left({A}+{X}\right)\left({B}+{X}\right)=−\mathrm{1} \\ $$$${X}^{\mathrm{2}} +\left({A}+{B}\right){X}+\left({AB}+\mathrm{1}\right)=\mathrm{0} \\ $$$${X}=\frac{−\left({A}+{B}\right)\pm\sqrt{\left({A}−{B}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{log}\:{x}=\frac{−\left(\mathrm{log}\:{a}+\mathrm{log}\:{b}\right)\pm\sqrt{\left(\mathrm{log}\:{a}−\mathrm{log}\:{b}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{log}\:{x}=\frac{−\mathrm{log}\:{ab}\pm\sqrt{\left(\mathrm{log}\:\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{10}^{\frac{−\mathrm{log}\:{ab}\pm\sqrt{\left(\mathrm{log}\:\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}} \\ $$$$ \\ $$$$\left(\mathrm{log}\:\frac{{a}}{{b}}\right)^{\mathrm{2}} \geqslant\mathrm{4} \\ $$$$\mathrm{log}\:\frac{{a}}{{b}}\leqslant−\mathrm{2}\:{or}\:\geqslant\mathrm{2} \\ $$$$\Rightarrow\:\mathrm{0}<\frac{{a}}{{b}}\leqslant\frac{\mathrm{1}}{\mathrm{100}}\:{or}\:\frac{{a}}{{b}}\geqslant\mathrm{100} \\ $$
Commented by jagoll last updated on 07/Feb/20
$${thank}\:{you}\:{mister}\:{W} \\ $$