logarithm question help needed please!?

it is already simplified

to factor it:

xln(x) − 8 + 2ln(x) − 4x = 0

=> xln(x) + 2ln(x) − 4x − 8 = 0

=> ln(x)(x + 2) − 4(x + 2) = 0

=> (x + 2)(ln(x) − 4) = 0

to solve it via factoring

x + 2 = 0 or ln(x) − 4 = 0

=> x = -2 or ln(x) = 4

=> x = -2 or e^4

given x = -2 will result in a log of a number 0 or less in the original equation

=> x = e^4

edit: @Learner actually the natural log of negative numbers is defined. because e^iπ = -1

=> ln(e^iπ) = ln(-1)

given log_a(a^n) = n

=> iπ = ln(-1)

and ln(-n) = ln(-1) + ln(n)

it you just need to be working in a context where complex numbers are accessible not just real numbers.

So it not correct to say that "the logs of negative numbers are undefined" [like you would say for dividing by zero] but you say "there are no real solutions to logs of negative numbers."

i) Grouping and factoring, (x + 2)*ln(x) - 4*(x + 2) = 0

==> (x + 2)*{ln(x) - 4} = 0

ii) So of the above, either x + 2 = 0 or ln(x) - 4 = 0

When x + 2 = 0, x = -2

and when ln|x| - 4 = 0, ln|x| = 4; ==> x = e⁴

But at x = -, ln(x) is not defined; hence this value is discarded.

Thus x = e⁴ only.

x ----------- x ----------- x ---------------- x ------------- x -------------- x --------- x

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xln(x)−8+2ln(x)−4x=0

xln(x)+2ln(x)-4x-8=0

ln(x)(x+2)-4(x+2)=0

(ln(x)-4)(x+2)=0

either ln(x)-4=0 => ln(x)=4 =>r x= e^4

or x+2=0 => x = -2

x.Ln(x) - 8 + 2.Ln(x) - 4x = 0 → where x > 0

x.Ln(x) + 2.Ln(x) - 4x - 8 = 0

(x + 2).Ln(x) - (4x + 8) = 0

(x + 2).Ln(x) - 4.(x + 2) = 0

(x + 2).[Ln(x) - 4] = 0

First case: (x + 2) = 0 → x + 2 = 0 → x = - 2 ← no possible because the condition

Second case: [Ln(x) - 4] = 0→ Ln(x) = 4 → x = e^4 ← possible