How do you solve #\log _ { 2} ( x + 2) + \log _ { 2} ( x - 1) = 2#?

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Answer 1 · 2017-01-03T19:16:56

#x = 1.5616#

When you work with logs with an equation, all the terms must be log terms before you can solve it.

#log_2(x+2)+log_2(x-1) = 2# can be written as:

#log_2(x+2)+log_2(x-1) = log_2 4" "larr log_2 4 = 2#

"Adding logs means multiply the terms"

#log_2 (x+2)(x-1) = log_2 4#

[#log_a B = log_a C " "hArr" " B =C# ]

#:. (x+2)(x-1) = 2#

#x^2 -x +2x-2 = 2#

#x^2 +x-4 =0#

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#x = (-1+-sqrt(1^2 -4(1)(-4)))/(2(1))#

#x = (-1+-sqrt(1+16))/2#

#x = (-1+sqrt(17))/2 " or "(-1-sqrt(17))/2#

#x =1.5616" or "x = -2.5616#

However log of a negative number is not defined, so

#x = 1.5616#