How do you solve # lnx+ln(x+2)=1#?

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Answer 1 · 2016-07-03T01:01:33

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

#ln(x(x + 2) = 1#

#ln(x^2 + 2x) = 1#

#x^2 + 2x = e^1#

#x^2 + 2x - e = 0#

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

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

#x = (-2 +- sqrt(4 + 4e))/2#

#x = 0.93 and -2.93#

However, checking the solutions in the original equation we find that #x= -2.93# is extraneous.

Hence, the solution set is #{0.93}#.

Hopefully this helps!