Question Number 4477 by love math last updated on 30/Jan/16
$$\mathrm{2}\:{log}\:\left({x}−\mathrm{2}\right)+{log}\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$$${Determine}\:{the}\:{domain}\:{of}\:{x}\:{and}\:{find}\:{the}\:{value}\:{of}\:{x}. \\ $$
Answered by Rasheed Soomro last updated on 30/Jan/16
$$\mathrm{2}\:{log}\:\left({x}−\mathrm{2}\right)+{log}\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{log}\left({x}−\mathrm{2}\right)=−{log}\left({x}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$${log}\left({x}−\mathrm{2}\right)^{\mathrm{2}} ={log}\left({x}−\mathrm{4}\right)^{−\mathrm{2}} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\left({x}−\mathrm{4}\right)^{−\mathrm{2}} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{{x}−\mathrm{4}}\right)^{\mathrm{2}} \\ $$$${x}−\mathrm{2}=\pm\frac{\mathrm{1}}{{x}−\mathrm{4}} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{4}\right)=\pm\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}−\mathrm{1}=\mathrm{0}\:\:\mid\:{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}+\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7}=\mathrm{0}\:\:\:\mid\:\:{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}=\mathrm{0} \\ $$$${x}=\frac{−\left(−\mathrm{6}\right)\pm\sqrt{\left(−\mathrm{6}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{7}\right)}}{\mathrm{2}}\:\:\mid\:\:\left({x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{\mathrm{6}\pm\sqrt{\mathrm{8}}}{\mathrm{2}}=\mathrm{3}\pm\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\mid\:\:\:{x}=\mathrm{3} \\ $$$${Domain}\:{of}\:{equation}\::\left\{\mathrm{3},\mathrm{3}\pm\sqrt{\mathrm{2}}\right\} \\ $$$$ \\ $$