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Question-21342




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

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