Question
Loga^2+loga=1.solve for a.
Answers
oobleck
2loga + loga = 1
3loga = 1
loga = 1/3
a = 10^(1/3)
Or maybe you meant
log^2(a) + loga = 1
log^2(a) + loga - 1 = 0
loga = (-1±√5)/2
a = 10^((-1+√5)/2)
3loga = 1
loga = 1/3
a = 10^(1/3)
Or maybe you meant
log^2(a) + loga = 1
log^2(a) + loga - 1 = 0
loga = (-1±√5)/2
a = 10^((-1+√5)/2)
Anonymous
Let say; loga=y.then
2y+y=1
2y+y=1
3y=1.then
y=1/3.but
loga=1/3
a=10^1/3.
2y+y=1
2y+y=1
3y=1.then
y=1/3.but
loga=1/3
a=10^1/3.