Question 930565
if you have this:

{{{log(10,e^5)+log(10,(5x+1))=log(10,(x+5))+1}}} then {{{x}}} will be:

{{{log(10,e^5)+log(10,(5x+1))=log(10,(x+5))+log(10,10)}}}

{{{log(10,e^5(5x+1))=log(10,10(x+5))}}} ...since log same,

{{{e^5(5x+1)=10(x+5)}}} is same too

{{{e^5(5x+1)=10(x+5)}}}

{{{5xe^5+e^5=10x+50}}}

{{{5xe^5-10x=50-e^5}}}

{{{(5e^5-10)x=50-e^5}}}

{{{x=-(e^5-50)/(5e^5-10)}}}

{{{x = -(e^5-50)/5(e^5-2)}}}........exact solution

or,you can substitute => {{{e^5=148.4}}} and get approximate solution


{{{148.4(5x+1)=10(x+5)}}}

{{{742x+148.4=10x+50}}}

{{{742x-10x=50-148.4}}}

{{{732x=-98.4}}}

{{{x=-98.4/732}}}

{{{x=-0.1344262295081967)}}}


if you plug in this solution in {{{log(10,e^5)+log(10,(5x+1))=log(10,(x+5))+1}}} and you check it, you will get {{{1.687172570169473=1.68713406159622243}}} which could be rounded to  {{{1.6871=1.6871}}} to be same (error is to small)