Question 118903
{{{log(2,(3x+2))-log(4,(x))=3}}} Start with the given equation




{{{log(10,(3x+2))/log(10,(2))-log(10,(x))/log(10,(4))=3}}} Use the change of base formula {{{log(b,x)=log(10,x)/log(10,b)}}} to rewrite every log in base 10



{{{log(10,(3x+2))/log(10,(2))-log(10,(x))/log(10,(4))=3}}} Rewrite {{{log(10,(4))}}} as {{{log(10,2^2)}}}



{{{log(10,(3x+2))/log(10,(2))-log(10,(x))/2log(10,(2))=3}}} Rewrite {{{log(10,2^2)}}} as {{{2log(10,(2))}}}




{{{2log(10,(3x+2))/2log(10,(2))-log(10,(x))/2log(10,(2))=3}}} Multiply the first fraction by {{{2/2}}}




{{{(2log(10,(3x+2))-log(10,(x)))/2log(10,(2))=3}}} Combine the fractions




{{{(log(10,(3x+2)^2)-log(10,(x)))/2log(10,(2))=3}}} Rewrite {{{2log(10,(3x+2))}}} as {{{log(10,(3x+2)^2)}}}




{{{(log(10,(3x+2)^2/x))/2log(10,(2))=3}}} Combine the logs




{{{(log(10,(3x+2)^2/x))/log(10,(4))=3}}} Rewrite {{{2log(10,(2))}}} as {{{log(10,(4))}}}




{{{(log(4,(3x+2)^2/x))=3}}} Use the change of base formula again to rewrite the log




{{{4^3=(3x+2)^2/x}}} Now use the property {{{log(b,x)=y}}} ---> {{{b^y=x}}}



{{{64=(3x+2)^2/x}}} Raise 4 to the 3rd power to get 64



{{{64x=(3x+2)^2}}} Multiply both sides by x



{{{64x=9x^2+12x+4}}} Foil



{{{0=9x^2-52x+4}}} Subtract 64x from both sides




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 {{{9*x^2-52*x+4=0}}} ( notice {{{a=9}}}, {{{b=-52}}}, and {{{c=4}}})





{{{x = (--52 +- sqrt( (-52)^2-4*9*4 ))/(2*9)}}} Plug in a=9, b=-52, and c=4




{{{x = (52 +- sqrt( (-52)^2-4*9*4 ))/(2*9)}}} Negate -52 to get 52




{{{x = (52 +- sqrt( 2704-4*9*4 ))/(2*9)}}} Square -52 to get 2704  (note: remember when you square -52, you must square the negative as well. This is because {{{(-52)^2=-52*-52=2704}}}.)




{{{x = (52 +- sqrt( 2704+-144 ))/(2*9)}}} Multiply {{{-4*4*9}}} to get {{{-144}}}




{{{x = (52 +- sqrt( 2560 ))/(2*9)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (52 +- 16*sqrt(10))/(2*9)}}} 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 = (52 +- 16*sqrt(10))/18}}} Multiply 2 and 9 to get 18


So now the expression breaks down into two parts


{{{x = (52 + 16*sqrt(10))/18}}} or {{{x = (52 - 16*sqrt(10))/18}}}



Now break up the fraction



{{{x=+52/18+16*sqrt(10)/18}}} or {{{x=+52/18-16*sqrt(10)/18}}}



Simplify



{{{x=26 / 9+8*sqrt(10)/9}}} or {{{x=26 / 9-8*sqrt(10)/9}}}



So these expressions approximate to


{{{x=5.69980236459412}}} or {{{x=0.0779754131836626}}}



So our solutions are:

{{{x=5.69980236459412}}} or {{{x=0.0779754131836626}}}