Answer: n = 3
Step-by-step explanation:
We will solve the given natural log equation for n.
Given:
[tex]ln(-3+3n)=ln(n^{2}-n )[/tex]
Raise e to the power of both sides of the equation:
[tex]\displaystyle e^{ln(-3+3n)}=e^{ln(n^{2}-n )}[/tex]
[tex]\displaystyle -3+3n=n^{2}-n[/tex]
Subtract (n² - n) from both sides of the equation:
[tex]\displaystyle -3+n+3n-n^{2}=0[/tex]
Combine like terms:
[tex]\displaystyle -3+4n-n^{2}=0[/tex]
Reorder in standard form:
[tex]\displaystyle -n^{2}+4n-3=0[/tex]
Factor out a -1:
[tex]\displaystyle -(n^{2}-4n+3)=0[/tex]
Divide both sides of the equation by -1:
[tex]\displaystyle n^{2}-4n+3=0[/tex]
Factor the expression n² - 4n + 3:
[tex]\displaystyle (n-3)(n-1)=0[/tex]
Apply the zero product property:
[tex]\displaystyle n=1,3[/tex]
Confirm if both solutions apply:
[tex]ln(-3+3(1))=ln((1)^{2}-(1) )\;\rightarrow \; ln(-3+3)=ln(1-1) \; \rightarrow \text{n=1 is und}\text{efined}[/tex]
[tex]ln(-3+3(3))=ln((3)^{2}-(3) )\;\rightarrow \; ln(-3+9)=ln(9-3)\; \rightarrow n=3 \;\checkmark[/tex]
Solution:
[tex]\displaystyle n=1,\boxed{3}[/tex]
n = 3