if log 2 to the base x+ log 8 the base x+ log 64 to the base xequal to 3 then x

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Melody · Answer

large underline{ sf{Solution-}}     Given that      rm : longmapsto : log_{x}(2) +  log_{x}(8) +  log_{x}(64) = 3        rm  :   :   rm : longmapsto : :   :  log_{x}(2  times 8  times 64)  = 3        :   :   :   :   :  red{ bigg  {  because  : log_{a}(x) +  log_{a}(y)=log_{a}(xy) bigg  }}       rm  :   :   rm : longmapsto :   :   :  log_{x}(2  times  {2}^{3}  times  {2}^{6})  = 3       rm  :   :   rm : longmapsto :  :   :  log_{x}( {2}^{1 + 3 + 6} )  = 3        :   :   :   :   :   :   :   :   :   :   :   :   :   :  red{ bigg  {  because  :  {a}^{x}   times  {a}^{y}  =  {a}^{x + y}  bigg  }}       rm  :   :   rm : longmapsto :   :   :  log_{x}( {2}^{10} )  = 3       rm : longmapsto : {2}^{10} =  {x}^{3}         :   :   :   :   :   :   :   :   :   :   :  red{ bigg  {  because  : log_{x}(y) = z  : then  : y =  {x}^{z}    bigg  }}       rm : longmapsto : {x}^{3}  =  {2}^{9}   times 2       bf implies  :x =  {2}^{3}  times  {2}^{  frac{1}{3} }        bf implies  :x = 8  :   sqrt[3]{2}         Additional Information  :-         boxed{  sf  : log_{x}(x)  = 1 }       boxed{  sf  : log_{x}(y)  =  dfrac{1}{ log_{y}(x) }  }       boxed{  sf  :  {a}^{ log_{a}(x) } = x }       boxed{  sf  :  {a}^{ ylog_{a}(x) } =  {x}^{y}  }       boxed{  sf  :  log(1) = 0 }       boxed{  sf  :  {e}^{ log_{e}(x) }  = x}

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