alrighty
remember
[tex]log_a(b)=c[/tex] means [tex]a^c=b[/tex]
and
[tex]ln(x)=log_e(x)[/tex]
and
[tex]x^0=1[/tex] for all real values of x
so
[tex]ln(x^2+4x-5)=0[/tex] means
[tex]log_e(x^2+4x-5)=0[/tex] means
[tex]e^0=x^2+4x-5[/tex] which simplifies to
1=x²+4x-5
minus 1 both sides
0=x²+4x-6
use quadratic formula or complete the square
[tex]x=-2+\sqrt{10}[/tex] and [tex]x=-2-\sqrt{10}[/tex]
2 solutions
