Question 877781
use log definition {{{a=ln(e^a)}}}
{{{6=ln(e^6)}}} , {{{13=ln(e^13)}}}
{{{ln(2x+1) +ln(e^6)=ln(e^13)}}}

for logs with the same base, use definition {{{log(b,(f(x)))+log(b,(g(x)))=log(b,(f(x))*(g(x)))}}}

{{{ln((2x+1)e^6)=ln(e^13)}}}

when logs have the same base {{{log(b,(f(x)))=log(b,(g(x)))}}} where {{{f(x)=g(x)}}}
 so
{{{(2x+1)(e^6)=(e^13)}}}
now solve for x

{{{(2x+1)(e^6)=(e^13)}}}
{{{2x(e^6) +(e^6)=(e^13)}}}
{{{2x(e^6)=(e^13)-(e^6)}}}
{{{2x=((e^13)-(e^6))/(e^6)}}}
{{{x=((e^13)-(e^6))/2(e^6)}}}

verify:
substitute {{{x=((e^13)-(e^6))/2(e^6)}}} into original equation {{{ln(2x+1)+6=13 }}}

{{{ln(2((e^13)-(e^6))/2(e^6)+1)+6=13)}}} 
={{{ (ln(2(548.8165))+6)=13) }}}
={{{(ln(1097.633))+6=13}}}
={{{7.000+6=13}}}
={{{13=13}}}