Question 917068


{{{ log((2x+1))-log((3x-2))=1 }}}


{{{log((2x+1)/(3x-2))=1}}} ...in base {{{10}}} we will have


{{{log((2x+1)/(3x-2))/log(10)=1}}}


{{{log((2x+1)/(3x-2))=log(10)}}} ....if log same, then


{{{(2x+1)/(3x-2)=10}}}


{{{(2x+1)=10(3x-2)}}}


{{{2x+1=30x-20}}}


{{{20+1=30x-2x}}}


{{{21=28x}}}


{{{21/28=x}}}


{{{3/4=x}}}