1.) e^(2x) - 3e^(x) - 4 = 0
2.) e^(1-2x) ((e^(3x+6) - e^(2x)) = 0
The first one is a
The first one is a quadratic...
(e^x-4)(e^x+1) you need to spot these, You will never see one in real life engineering, but they are put in math books by the pleura.
The second one I would multiply it thru
e^(x+7) -e which = e(e^(x+6) -1)
To solve the given equations, we need to isolate the variable "x" on one side of the equation. Let's solve them step by step:
1.) e^(2x) - 3e^(x) - 4 = 0
Step 1: Let's make a substitution. Set u = e^x.
The equation becomes: u^2 - 3u - 4 = 0.
Step 2: Factor the quadratic equation.
(u - 4)(u + 1) = 0.
Step 3: Solve for u.
Set each factor equal to zero:
u - 4 = 0, which gives u = 4.
u + 1 = 0, which gives u = -1.
Step 4: Replace u with e^x.
For u = 4: e^x = 4 > x = ln(4).
For u = -1: e^x = -1 (not possible since e^x is always positive).
Therefore, the solution is x = ln(4).
2.) e^(1-2x) ((e^(3x+6) - e^(2x)) = 0
Step 1: Apply exponent rules.
Distribute e^(1-2x) to the terms inside the parentheses:
e^(1-2x) * e^(3x+6) - e^(1-2x) * e^(2x) = 0.
Step 2: Simplify the exponents.
e^(1-2x+3x+6) - e^(1-2x+2x) = 0.
e^(4+x) - e^(1-x) = 0.
Step 3: Combine like terms.
e^(4+x) - e^(1-x) = 0.
To further solve this equation, we need more information or additional equations since it cannot be factored or simplified any further.