Question 186944
{{{2*ln(x)+ln(x+7)=3*ln(x+1)}}} Start with the given equation.



Note: the domain here is {{{x>0}}} (since we CANNOT plug in zero or negative values into {{{2*ln(x)}}})



{{{ln(x^2)+ln(x+7)=ln((x+1)^3)}}} Rewrite the logs using the identity  {{{y*ln(x)=ln(x^y)}}} (ie move the numbers out front the logs to the exponent position)



{{{ln(x^2(x+7))=ln((x+1)^3)}}} Combine the logs on the left side using the identity {{{ln(A)+ln(B)=ln(A*B)}}}



{{{x^2(x+7)=(x+1)^3}}} Raise both sides as exponents with the base of 'e' (this effectively cancels out the natural logs)



{{{x^2(x+7)=(x+1)(x+1)(x+1)}}} Rewrite {{{(x+1)^3}}} as {{{(x+1)(x+1)(x+1)}}}



{{{x^2(x+7)=(x+1)(x^2+2x+1)}}} FOIL the last two binomials



{{{x^2(x+7)=x(x^2+2x+1)+1(x^2+2x+1)}}} Expand



{{{x^2(x+7)=x^3+2x^2+x+x^2+2x+1}}} Distribute



{{{x^3+7x^2=x^3+3x^2+3x+1}}} Combine like terms.



{{{x^3+7x^2-x^3-3x^2-3x-1=0}}} Get everything to the left side.



{{{4x^2-3x-1=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=-3}}}, and {{{c=-1}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-3) +- sqrt( (-3)^2-4(4)(-1) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=-3}}}, and {{{c=-1}}}



{{{x = (3 +- sqrt( (-3)^2-4(4)(-1) ))/(2(4))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(4)(-1) ))/(2(4))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--16 ))/(2(4))}}} Multiply {{{4(4)(-1)}}} to get {{{-16}}}



{{{x = (3 +- sqrt( 9+16 ))/(2(4))}}} Rewrite {{{sqrt(9--16)}}} as {{{sqrt(9+16)}}}



{{{x = (3 +- sqrt( 25 ))/(2(4))}}} Add {{{9}}} to {{{16}}} to get {{{25}}}



{{{x = (3 +- sqrt( 25 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (3 +- 5)/(8)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{x = (3 + 5)/(8)}}} or {{{x = (3 - 5)/(8)}}} Break up the expression. 



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



{{{x = 1}}} or {{{x = -1/4}}} Simplify. 



So the possible answers are {{{x = 1}}} or {{{x = -1/4}}} 



However, since you CANNOT take the natural of a negative number, this means that {{{x = -1/4}}} is NOT a solution. 



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Answer:


So the solution is {{{x = 1}}}