Question 448342
First use the idea that {{{log(b,(A))-log(b,(B))=log(b,(A/B))}}}



So the left side should condense into {{{log(4,((x^2+3x)/(x+5)))}}} where in this case {{{A=x^2+3x}}}, {{{B=x+5}}} and {{{b=4}}}




So the equation simplifies to {{{log(4,((x^2+3x)/(x+5)))=1}}}



Now you can use the property you were referring to and you'll get


{{{4^1=(x^2+3x)/(x+5)}}}



{{{4=(x^2+3x)/(x+5)}}} Simplify



{{{4(x+5)=x^2+3x}}} Multiply both sides by {{{x+5}}}



{{{4x+20=x^2+3x}}} Distribute.



{{{0=x^2+3x-4x-20}}} Get everything to the right side.



{{{0=x^2-x-20}}} Combine like terms.



Notice that the quadratic {{{x^2-x-20}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=-20}}}



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



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-1) +- sqrt( (-1)^2-4(1)(-20) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=-20}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(-20) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(-20) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--80 ))/(2(1))}}} Multiply {{{4(1)(-20)}}} to get {{{-80}}}



{{{x = (1 +- sqrt( 1+80 ))/(2(1))}}} Rewrite {{{sqrt(1--80)}}} as {{{sqrt(1+80)}}}



{{{x = (1 +- sqrt( 81 ))/(2(1))}}} Add {{{1}}} to {{{80}}} to get {{{81}}}



{{{x = (1 +- sqrt( 81 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- 9)/(2)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{x = (1 + 9)/(2)}}} or {{{x = (1 - 9)/(2)}}} Break up the expression. 



{{{x = (10)/(2)}}} or {{{x =  (-8)/(2)}}} Combine like terms. 



{{{x = 5}}} or {{{x = -4}}} Simplify. 



So the <u>possible</u> solutions are {{{x = 5}}} or {{{x = -4}}} 



However, we need to check them to verify that they are actually solutions. To do so, just plug in the values of x and simplify to get an identity. 



It turns out that both solutions check out. So the two solutions are {{{x = 5}}} or {{{x = -4}}}