Question Number 6111 by Kasih last updated on 14/Jun/16
$${log}^{\mathrm{2}} {x}+{logx}^{\mathrm{2}} −\mathrm{3}=\mathrm{0}\:,\:{x}=? \\ $$
Answered by prakash jain last updated on 14/Jun/16
$$\mathrm{log}\:{x}={u} \\ $$$${u}^{\mathrm{2}} +\mathrm{2}{u}−\mathrm{3}=\mathrm{0}\Rightarrow\left({u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=−\mathrm{3}\:{or}\:{u}=\mathrm{1} \\ $$$$\mathrm{log}\:{x}=−\mathrm{3}\:{or}\:\mathrm{log}\:{x}=\mathrm{1} \\ $$$${x}=\mathrm{10}^{−\mathrm{3}} \:{or}\:{x}=\mathrm{10}^{\mathrm{1}} \\ $$