Question 792472
{{{ log( sqrt (x), 3 ) +  log( x^2, 16 ) =1 }}} 
Expressing everything using base 10 logarithms, we get
{{{log(3)/log(sqrt(x))+log(16)/log((x^2))=1}}} ---> {{{log(3)/log((x^0.5))+log(16)/log((x^2))=1}}} ---> {{{log(3)/(0.5*log(x))+log(16 )/(2*log(x))=1}}} --->
 
{{{2*log( 3 )/log (x) + 0.5log( 16 )/log( x) =1 }}} ---> {{{(2*log( 3 )+ 0.5*log( 16 ))/log( x) =1 }}} ---> {{{(log((3^2))+ log((16^0.5)))/log( x) =1 }}} --->
 
{{{(log(9)+log(4))/log(x)=1}}} ---> {{{log((9*4))/log(x)=1}}} ---> {{{log(36)/log(x) =1 }}} --->
 
{{{log(36)=log(x)}}} --> {{{36=x}}}