Question Number 21342 by xxyy@gmail.com last updated on 21/Sep/17
Answered by $@ty@m last updated on 23/Sep/17
$${Let}\:\mathrm{10}+\mathrm{log}_{\mathrm{10}} \left({x}\right)^{\mathrm{10}+\mathrm{log}_{\mathrm{10}} \left({x}\right)^{\mathrm{10}+….} } ={y} \\ $$$$\Rightarrow\mathrm{10}+\mathrm{log}\:_{\mathrm{10}} {x}^{{y}} ={y}\:\:−−\left(\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{10}} {x}^{{y}} ={y}−\mathrm{10} \\ $$$$\Rightarrow{y}\mathrm{log}\:_{\mathrm{10}} {x}={y}−\mathrm{10} \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{10}} {x}=\frac{{y}−\mathrm{10}}{{y}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} {x}}=\frac{{y}}{{y}−\mathrm{10}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} {x}}=\frac{\mathrm{10}+\mathrm{log}\:{x}^{{y}} }{\mathrm{log}\:{x}^{{y}} }\:,\:{using}\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} {x}}=\mathrm{1}+\frac{\mathrm{10}}{\mathrm{log}\:{x}^{{y}} }\:\:−−\left(\mathrm{2}\right) \\ $$$${ATQ}, \\ $$$$\mathrm{1}+{x}^{{y}} =\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} {x}}\:−−\left(\mathrm{3}\right) \\ $$$${from}\:\left(\mathrm{2}\right)\:\&\:\left(\mathrm{3}\right), \\ $$$${x}^{{y}} =\frac{\mathrm{10}}{\mathrm{log}{x}^{{y}} } \\ $$$$\mathrm{log}\:\left({x}^{{y}} \right)^{{x}^{{y}} } =\mathrm{10} \\ $$$$\left({x}^{{y}} \right)^{{x}^{{y}} } =\mathrm{10}^{\mathrm{10}} \\ $$$${x}^{{y}} =\mathrm{10} \\ $$$$\mathrm{1}+{x}^{{y}} =\mathrm{11} \\ $$$$\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{10}} {x}}=\mathrm{11},\:{from}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{log}_{\mathrm{10}} {x}=\frac{\mathrm{1}}{\mathrm{11}} \\ $$$${x}=\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{11}}} \:{Ans}. \\ $$$$ \\ $$
Commented by xxyy@gmail.com last updated on 23/Sep/17
$$\mathrm{thank}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$