How do you solve #Log(x+2)+log(x-1)=1 #?

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Answer 1 · 2016-02-29T01:38:28

Use the log rule #log_am + log_an = log_a(m xx n)#

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

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

#log(x^2 + 2x - x - 2) = 1#

#log(x^2 + x - 2) = 1#

Convert to exponential form.

#x^2 + x - 2 = 10^1#

#x^2 + x - 12 = 0#

#(x + 4)(x - 3) = 0#

#x = -4 and 3#

Since the log of a negative number is undefined the solution is x = 3.

Hopefully this helps.