How do you solve #\log _ { 3} ( x - 3) + \log _ { 3} ( x + 4) = \log _ { 3} 8#?

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Answer 1 · 2017-07-28T22:03:14

#x = 4#

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#

#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#.