Solve the natural log for x

Question

ln(x+2) + lnx = ln(x+12)

Answer 1

ln(x+2) + ln(x) = ln(x+12)

ln(x(x+2)) = ln(x+12) <-- Note that ln(a) + ln(b) = ln(ab)

ln(x²+2x) = ln(x+12)

x²+2x = x+12

x²+x = 12

x²+x-12 = 0

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

x = -4,3

Going back to the original equation, ln(x+2) + ln(x) = ln(x+12), we see that if we plug in x=-4, then we run into ln(-4) on the right-hand side, which is not real. Therefore, only x=3 is the correct answer.

Hope this helped!

Answer 2

logx + log(x+2) = log(x+12)

log(x^2 +2x) = log(x+12)

x^2 +x -12 = 0

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

x =-4,3

log3 + log5 =log(3x5) = log15 = log(3+12)

x=3 works as a valid solution

but whenever you multiply by a variable at any stage of the problem you risk possibly introducing an extraneous solution that is not in the original problem

check -4

log(-4)+log(-2) = log(8) = log(-4+12)

it also seems to work

but

logarithms generally have a domain of only non-negative numbers

logs of negative numbers can be imaginary though, and imaginary solutions are mathematically recognized

consider Euler's Identity

e^ipi = -1 = cospi +isinpi=-1+0

but

e^i(-ip) also =-1 = cos(-pi) +isin(-pi) = -1+0

e^ipi = cospi +isinpi

take logs of both sides

ipi = ln(cospi+isinpi) = ln(-1+0)

ln(-1) = ipi

e^i(x) = cosx +isinx

take logs of both sides

ix = log(cosx) +isinx)

if x= pi (or -pi, a negative value for taking logs, but same solution as for x=+pi)

x= log(cospi) + log(isinpi)

log(-1) =-ipi or ipi

but do you get even an imaginary solution when x=-4 and is -4 referring to radians or degrees

just some "minor" complications in considering two possible solutions.

even if log-4 isn't imaginary, it could be another category of non-real numbers