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..
tg(x)sec^6(x)dx
To solve the logarithm problem, start by letting y = ln(x) be a new variable. This substitution helps simplify the problem.
Now, we have the equation 2y^2 + y - 1 = 0.
To solve this quadratic equation, we can use the factoring method or the quadratic formula.
Let's use the factoring method. We want to find two numbers that multiply to give -2 (-1 * 2) and add up to 1.
After trying some combinations, we can see that -1 and 1/2 satisfy this condition. Therefore, we can factor the equation as follows:
2(y + 1)(y - 1/2) = 0.
Using the zero-product property, we can set each factor equal to zero:
y + 1 = 0 --> y = -1
y - 1/2 = 0 --> y = 1/2
Now, we have two possible values for y: -1 and 1/2.
To find the corresponding values of x, we use the fact that x = e^y (where e is the base of the natural logarithm).
For y = -1:
x = e^(-1) = 1/e ≈ 0.3679
For y = 1/2:
x = e^(1/2) ≈ 1.6487
Therefore, the solutions for the equation 2(ln(x))^2 + lnx - 1 = 0 are approximately x ≈ 0.3679 and x ≈ 1.6487.