(ex)x·e4x + 4 = 1… so i need to solve for x…. how the hell do I do that???

Hey sorry the equation is a little different than that…

(ex)x·e4x + 4 = 1

all the xs and 4x+4 are all exponents.

If you mean e^x * e^(4x+4) = 1

just take log of both sides:

x + (4x+4) = 0
5x = -4
x = -4/5

check

e^(-4/5) * e^(-16/5 + 20/5) = e^(-4/5) * e^(4/5) = e^(-4/5 + 4/5) = e^0 = 1

To solve the equation x·e^(4x) + 4 = 1, you need to find the value(s) of x that satisfy the equation. Here's a step-by-step process to solve it:

Step 1: Subtract 4 from both sides of the equation:
x·e^(4x) = -3

Step 2: Take the natural logarithm (ln) of both sides of the equation:
ln(x·e^(4x)) = ln(-3)

Step 3: Use the logarithmic property to simplify the left side of the equation:
ln(x) + ln(e^(4x)) = ln(-3)
ln(x) + 4x = ln(-3)

Step 4: Rearrange the equation to isolate the logarithmic term:
ln(x) = -4x + ln(-3)

Step 5: Exponentiate both sides of the equation using the property of exponentiation:
e^(ln(x)) = e^(-4x + ln(-3))
x = e^(-4x + ln(-3))

Step 6: This equation cannot be solved algebraically to find an exact solution because of the presence of both x and exponentials. You will need to use numerical methods, such as iteration or approximation algorithms, to find an approximate solution.