Question 1087097
IF you have this:

{{{R=(2/3)log( E)-3 }}}

{{{R+3=(2/3)log( E) }}}

{{{(R+3)/(2/3)=log( E) }}}

{{{3(R+3)/2=log( E) }}}

{{{log( E)= 3(R+3)/2}}}.....since  {{{y = log(b,x)}}} is equivalent to {{{x = b^y}}}, and in your case  {{{y =(3R+9)/2}}}, base {{{b=10}}}, and {{{x=E}}}

so, you will have:


{{{highlight(E=10^(3(R+3)/2))}}}