Question 473030
{{{ln(3x+5)+ln(3x-5)=6+ln4}}}
{{{system(3x+5>0, 3x-5>0)}}} => {{{system(3x>-5,  3x>5)}}} => {{{system(x>-5/3,  x>5/3)}}} => {{{x>5/3}}}
Let write 6 in the  way {{{6=ln(e^6)}}}
{{{ln(3x+5)+ln(3x-5)=ln(e^6)+ln4}}}
Use formula {{{lna+lnb=ln(ab)}}}
{{{ln((3x+5)(3x-5))=ln(4e^6)}}}
{{{(3x+5)(3x-5)=4e^6}}}
Use formula {{{(a+b)(a-b)=a^2-b^2}}}
{{{9x^2-25=4e^6}}} add 25 to the both sides
{{{9x^2=4e^6+25}}} divide by 9
{{{x^2=(4e^6+25)/9}}}
{{{x1=sqrt((4e^6+25)/9)}}}
{{{x2=-sqrt((4e^6+25)/9)<5/3}}}extraneous root
Answer {{{x=sqrt((4e^6+25)/9)}}}