e^(-x/2) + xe^(-x/2) = 0
solve for x
To solve the equation e^(-x/2) + xe^(-x/2) = 0, we need to isolate x. Let's break down the steps to solve it:
Step 1: Factor out the common term e^(-x/2) from both terms:
e^(-x/2)(1 + x) = 0
Step 2: Set each factor equal to zero:
e^(-x/2) = 0 or (1 + x) = 0
Step 3: Solve each factor separately:
For e^(-x/2) = 0:
To find the natural logarithm of both sides, we take the ln (natural logarithm) of both sides of the equation:
ln(e^(-x/2)) = ln(0)
(-x/2) * ln(e) = ln(0)
Note that ln(e) equals 1 since the natural logarithm of e is 1.
So we have:
(-x/2) * 1 = ln(0)
At this point, we encounter a problem because ln(0) is undefined. Hence, there is no solution for e^(-x/2) = 0.
Moving on to the second factor, (1 + x) = 0:
Subtract 1 from both sides of the equation:
1 + x - 1 = 0 - 1
x = -1
So the solution to the equation e^(-x/2) + xe^(-x/2) = 0 is x = -1.