- #1
Robokapp
- 218
- 0
can anyone explain the behaviour of this function? i looked at the graph and everything, but if i solve it one way i get a different answer than if i solve it another way.
Ln[(x+1)^2]=2
one way to solve it (What i did)
e^{Ln[(x+1)^2]}=e^2
(x+1)^2=e^2
(x+1)^2-e^2=0
(x+1-e)(x+1+e)=0
x=e-1 and x=-e-1
another way, what my math teacher did
Ln[(x+1)^2]=2
2Ln(x+1)=2
Ln(x+1)=1
X+1=e^1
x=e-1
as you can see somehow one of his roots, the -e-1 is just...gone. i checked the math 1 million times. this is puzzling me for a few months now...where did the root go? i never asked him becasue he's knida cranky...but can you help me please?
Ln[(x+1)^2]=2
one way to solve it (What i did)
e^{Ln[(x+1)^2]}=e^2
(x+1)^2=e^2
(x+1)^2-e^2=0
(x+1-e)(x+1+e)=0
x=e-1 and x=-e-1
another way, what my math teacher did
Ln[(x+1)^2]=2
2Ln(x+1)=2
Ln(x+1)=1
X+1=e^1
x=e-1
as you can see somehow one of his roots, the -e-1 is just...gone. i checked the math 1 million times. this is puzzling me for a few months now...where did the root go? i never asked him becasue he's knida cranky...but can you help me please?