1. e^ln3 - 2lnx
2. 2(lnx)^2 + lnx - 1 =0
3. d/dx (e^x . e^e^x)
1.
e^ln 3 = 3
so
3 - ln x^2
2.
2z^2 +z-1 = 0
(2z-1)(z+1) = 0
z = 1/2 or -1
ln x = 1/2
e^ln x = x = e^.5 = 1.65
ln x = -1
e^-1 = 1/e = .367
3.
do not understand notation
1. To simplify the expression e^ln3 - 2lnx, we can use the properties of logarithms and exponentials. The natural logarithm, ln(x), is the inverse of the natural exponential function, e^x. Applying this relationship, we can simplify the expression as follows:
e^ln3 - 2lnx = 3 - 2lnx
2. To solve the equation 2(lnx)^2 + lnx - 1 = 0, we can use a technique called factoring. Let's break down the steps:
2(lnx)^2 + lnx - 1 = 0
Now, let's factor the left side of the equation:
(2lnx - 1)(lnx + 1) = 0
To find the values that make the equation true, we can set each factor equal to zero and solve for x:
2lnx - 1 = 0 -> lnx = 1/2 -> x = e^(1/2)
lnx + 1 = 0 -> lnx = -1 -> x = e^(-1)
Therefore, the solutions to the equation are x = e^(1/2) and x = e^(-1).
3. To differentiate the expression d/dx (e^x . e^e^x), we can use the product rule of differentiation. The product rule states that for two functions u(x) and v(x), the derivative of their product is given by:
d/dx(u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
Applying the product rule to our expression, we get:
d/dx (e^x . e^e^x) = d/dx (e^x) * e^e^x + e^x * d/dx (e^e^x)
The derivative of e^x is itself (since it is its own derivative), and the derivative of e^e^x can be found using the chain rule as e^e^x * d/dx(e^x), which simplifies to e^e^x * e^x. Therefore:
d/dx (e^x . e^e^x) = e^x * e^e^x + e^x * e^e^x = 2e^x * e^e^x
So, the derivative of (e^x . e^e^x) with respect to x is 2e^x * e^e^x.