Question 926058
{{{log(x+2)+log(x-3)=log(4x)}}}

{{{log((x+2)(x-3))=log(4x)}}}...since log same, we have

{{{(x+2)(x-3)=4x}}}...solve for {{{x}}}

{{{x^2-3x+2x-6=4x}}}

{{{x^2-x-6=4x}}}

{{{x^2-x-6-4x=0}}}

{{{x^2-5x-6=0}}}

{{{x^2+x-6x-6=0}}}

{{{(x^2+x)-(6x+6)=0}}}

{{{x(x+1)-6(x+1)=0}}}

{{{(x-6)(x+1)=0}}}

solutions:

if {{{(x-6)=0}}}=> {{{highlight(x=6)}}} (real solution)

if {{{(x+1)=0}}}=> {{{highlight(x=-1)}}}  (assuming a complex-valued logarithm)

intersection point: ({{{x}}},{{{y}}})

if {{{x=6}}}
{{{log(10,4x)=log(10,4*6)=log(10,24)=1.38}}}=> {{{y=1.38}}}

({{{6}}},{{{1.38}}})


{{{drawing( 600, 600, -10, 10, -5, 10,locate(6,1.38,p(6,1.38)),circle(6,1.38,.12), graph( 600, 600, -10, 10, -5, 10, log(10,(x+2)(x-3)), log(10,4x))) }}}