Question
Can somebody solve this logarithm problem step by step for me please:
2(lnx)² + lnx-1 = 0
First let y = ln x be a new variable, , and solve the polynomial
2 y^2 + y - 1 = 0
2(y +1)(y - 1/2) = 0
Solutions are
y = -1 or +1/2
Thus x = e^y
and x= 1/e = 0.3679.. OR
e^0.5 = 1.6487..
2(lnx)² + lnx-1 = 0
First let y = ln x be a new variable, , and solve the polynomial
2 y^2 + y - 1 = 0
2(y +1)(y - 1/2) = 0
Solutions are
y = -1 or +1/2
Thus x = e^y
and x= 1/e = 0.3679.. OR
e^0.5 = 1.6487..
Answers
Anonymous
tg(x)sec^6(x)dx