How do you solve #log_b(x - 1) + log_b(x + 2) = log_b(8 - 2x)#?

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Answer 1 · 2018-04-15T13:28:36

#x=2#

Notice that all the logs are of the same base!

First of all #log_b(x-1)+log_b(x+2)=log_b((x-1)(x+2))#

Therefore

#log_b((x-1)(x+2))=log_b(8-2x)#

We can remove the logs since both sides are logarithms of the same base.

#(x-1)(x+2)=8-2x#

This is a simple quadratic equation.

#x^2 +x -2=8-2x#
#x^2 +3x -10=0#

This can be solved by completing the square (or by using the quadratic formula)

#x^2 +3x =10#

#x^2 +3x +(3/2)^2= 10+(3/2)^2#

#(x +3/2)^2 = 10+(3/2)^2#

#x +3/2 = ± sqrt{10+(3/2)^2#

#x = ± sqrt{10+(3/2)^2} - 3/2#

#x = ± sqrt{10+9/4} - 3/2#

#x = ± sqrt{49/4} - 3/2#

#x = ± 7/2 - 3/2#

#x = (-3± 7)/2#

--

#x_1 = (7-3)/2 = 2#

#x_2 = (-7-3)/2 = -5#

However, since you can't take the logarithm of a negative number, x cannot be less than 0. Therefore #x=2# is the solution