Question Number 205457 by BaliramKumar last updated on 21/Mar/24
Answered by Rasheed.Sindhi last updated on 21/Mar/24
$${x}+\mathrm{log}_{\mathrm{15}} \left(\mathrm{1}+\mathrm{3}^{{x}} \right)={x}\mathrm{log}_{\mathrm{15}} \mathrm{5}+\mathrm{log}_{\mathrm{15}} \mathrm{12} \\ $$$$\mathrm{log}_{\mathrm{15}} \mathrm{15}^{{x}} +\mathrm{log}_{\mathrm{15}} \left(\mathrm{1}+\mathrm{3}^{{x}} \right)=\mathrm{log}_{\mathrm{15}} \mathrm{5}^{{x}} +\mathrm{log}_{\mathrm{15}} \mathrm{12} \\ $$$$\mathrm{log}_{\mathrm{15}} \left(\mathrm{15}^{{x}} ×\left(\mathrm{1}+\mathrm{3}^{{x}} \right)\right)=\mathrm{log}_{\mathrm{15}} \left(\mathrm{5}^{{x}} ×\mathrm{12}\right) \\ $$$$\mathrm{15}^{{x}} ×\left(\mathrm{1}+\mathrm{3}^{{x}} \right)=\mathrm{5}^{{x}} ×\mathrm{12} \\ $$$$\frac{\mathrm{15}^{{x}} +\mathrm{45}^{{x}} }{\mathrm{5}^{{x}} }=\mathrm{12} \\ $$$$\left(\frac{\mathrm{15}}{\mathrm{5}}\right)^{{x}} +\left(\frac{\mathrm{45}}{\mathrm{5}}\right)^{{x}} =\mathrm{12} \\ $$$$\mathrm{3}^{{x}} +\mathrm{9}^{{x}} =\mathrm{12} \\ $$$$\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} +\mathrm{3}^{{x}} −\mathrm{12}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{12}=\mathrm{0};\:{y}=\mathrm{3}^{{x}} \\ $$$$\left({y}+\mathrm{4}\right)\left({y}−\mathrm{3}\right)=\mathrm{0} \\ $$$${y}=−\mathrm{4},\mathrm{3} \\ $$$$\mathrm{3}^{{x}} =−\mathrm{4}\:\Rightarrow{x}\notin\mathbb{R} \\ $$$$\mathrm{3}^{{x}} =\mathrm{3}^{\mathrm{1}} \Rightarrow{x}=\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Mar/24
$${Last}\:{digit}\:{must}\:{be}\:\mathrm{0} \\ $$$${Last}\:{digit}\:{can}\:{be}\:{occupied}\:{in}\:\mathrm{1}\:{way}. \\ $$$${first}\:{digit}\:{can}\:{be}\:{occupied}\:{in}\:\mathrm{4}\:{ways}. \\ $$$${Second}\:{digit}\:{can}\:{be}\:{occupied}\:{in}\:\mathrm{3}\:{ways}. \\ $$$${Third}\:{digit}\:{can}\:{be}\:{occupied}\:{in}\:\mathrm{2}\:{ways}. \\ $$$${Number}\:{of}\:{ways}:\:\:\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{24} \\ $$$$ \\ $$
Commented by BaliramKumar last updated on 22/Mar/24
$$\mathrm{thanks} \\ $$
Answered by A5T last updated on 21/Mar/24
$${x}\left(\mathrm{1}−{log}_{\mathrm{15}} \mathrm{5}\right)={log}_{\mathrm{15}} \left(\frac{\mathrm{12}}{\mathrm{1}+\mathrm{3}^{{x}} }\right)\Rightarrow{x}=\frac{{log}_{\mathrm{15}} \left(\frac{\mathrm{12}}{\mathrm{1}+\mathrm{3}^{{x}} }\right)}{{log}_{\mathrm{15}} \mathrm{3}} \\ $$$$\Rightarrow{x}={log}_{\mathrm{3}} \left(\frac{\mathrm{12}}{\mathrm{1}+\mathrm{3}^{{x}} }\right)\Rightarrow\mathrm{3}^{{x}} \left(\mathrm{1}+\mathrm{3}^{{x}} \right)=\mathrm{12}\Rightarrow{p}^{\mathrm{2}} +{p}−\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow{p}=−\mathrm{4}\:{or}\:\mathrm{3}\Rightarrow{p}=\mathrm{3}^{{x}} =\mathrm{3}\Rightarrow{x}=\mathrm{1} \\ $$
Commented by BaliramKumar last updated on 22/Mar/24
$$\mathrm{thanks} \\ $$