We have: #log_(3)(x - 3) + log_(3)(x + 4) = log_(3)(8)#
Using the laws of logarithms:
#Rightarrow log_(3)((x - 3)(x + 4)) = log_(3)(8)#
#Rightarrow log_(3)((x - 3)(x + 4)) - log_(3)(8) = 0#
#Rightarrow log_(3)(frac((x - 3)(x + 4))(8)) = 0#
#Rightarrow frac((x - 3)(x + 4))(8) = 3^(0)#
#Rightarrow frac((x - 3)(x + 4))(8) = 1#
#Rightarrow (x - 3)(x + 4) = 8#
#Rightarrow x^(2) + 4 x - 3 x - 12 = 8#
#Rightarrow x^(2) + x - 20 = 0#
Let's use the quadratic formula:
#Rightarrow x = frac(- 1 pm sqrt(1^(2) - 4(1)(- 20))(2(1))#
#Rightarrow x = frac(- 1 pm sqrt(1 + 80))(2)#
#Rightarrow x = frac(- 1 pm sqrt(81))(2)#
#Rightarrow x = frac(- 1 pm 9)(2)#
#Rightarrow x = - 5, 4#
However, the arguments of logarithmic functions cannot be negative.
So #x = - 5# is not a solution.
Therefore, the solution to the equation is #x = 4#.
