Asked by Erik
I cant figure this question out. I could use some help.
Solve for x
lnx-lnx^2+ln5 = 0
Simplify the answer
Solve for x
lnx-lnx^2+ln5 = 0
Simplify the answer
Answers
Answered by
MathMate
Use the law of logarithms:
ln(a)+ln(b)=ln(ab)
ln(a)-ln(b)=ln(a/b)
kln(a)= ln(a<sup>k</sup>)
ln(1)=0
ln(1/x)=-ln(x)
so
ln(x)-ln(x²)+ln(5)=0
ln(x/x²)+ln(5)=0
ln(1/x)+ln(5)=0
-ln(x)+ln(5)=0
ln(x)=ln(5)
e^(ln(x))=e^(ln(5))
x=5
ln(a)+ln(b)=ln(ab)
ln(a)-ln(b)=ln(a/b)
kln(a)= ln(a<sup>k</sup>)
ln(1)=0
ln(1/x)=-ln(x)
so
ln(x)-ln(x²)+ln(5)=0
ln(x/x²)+ln(5)=0
ln(1/x)+ln(5)=0
-ln(x)+ln(5)=0
ln(x)=ln(5)
e^(ln(x))=e^(ln(5))
x=5