Question 139457
{{{log(5,(x^2))-log(5,(x+2)) = 2 }}} Start with the given equation



{{{log(5,x^2/(x+2)) = 2 }}} Combine the logs



{{{x^2/(x+2) = 5^2 }}} Use the relationship {{{log(b,(y))=x}}} <===> {{{b^x=y}}} to rewrite the equation



{{{x^2/(x+2) = 25 }}} Square 5



{{{x^2 = 25(x+2) }}} Multiply both sides by {{{x+2}}}



{{{x^2 = 25x+50 }}} Distribute



{{{x^2-25x-50=0  }}} Get all terms to one side



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{x^2-25*x-50=0}}} ( notice {{{a=1}}}, {{{b=-25}}}, and {{{c=-50}}})





{{{x = (--25 +- sqrt( (-25)^2-4*1*-50 ))/(2*1)}}} Plug in a=1, b=-25, and c=-50




{{{x = (25 +- sqrt( (-25)^2-4*1*-50 ))/(2*1)}}} Negate -25 to get 25




{{{x = (25 +- sqrt( 625-4*1*-50 ))/(2*1)}}} Square -25 to get 625  (note: remember when you square -25, you must square the negative as well. This is because {{{(-25)^2=-25*-25=625}}}.)




{{{x = (25 +- sqrt( 625+200 ))/(2*1)}}} Multiply {{{-4*-50*1}}} to get {{{200}}}




{{{x = (25 +- sqrt( 825 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (25 +- 5*sqrt(33))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (25 +- 5*sqrt(33))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (25 + 5*sqrt(33))/2}}} or {{{x = (25 - 5*sqrt(33))/2}}}





So these expressions approximate to


{{{x=26.861}}} or {{{x=-1.861}}}




So we can clearly see that {{{x=26.861}}} is the largest value of x that satisfies the equation