# What is x in equation 2/e^x = 5/(1+e^x)?

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We have to solve for x using the equation 2/e^x = 5/(1+e^x)

2/e^x = 5/(1+e^x)

=> 2* ( 1 + e^x) = 5* e^x

=> 2 + 2*e^x = 5*e^x

=> 2 = 3*e^x

=> (2/3) = e^x

take the natural logarithm of both the sides

=> x = ln (2/3)

=> x = -0.4054

Therefore x = ln (2/3) or -0.4054

To solve for x in equation 2/e^x = 5/(1+e^x).

We multiply the equation by the LCM e^x(1+e^x) of denominators and make e^x the subject:

2(1+e^x) = 5e^x.

2 = 5e^2-2e^x = 3e^x.

=> 3e^x = 2.

=> e^x = 2/3.

We take logarithms of both sides with respect to base e.:

x = ln(2/3).

Therefore x= ln(2/3).

We'll cross multiply:

2(1 + e^x) = 5e^x

We'll remove the brackets:

2 + 2e^x = 5e^x

We'll subtract 2e^x both sides:

2 = 5e^x - 2e^x

We'll use the symmetric property:

3e^x = 2

We'll divide by 3:

e^x = 2/3

We'll take logarithms both sides:

ln e^x = ln (2/3)

x* ln e = ln (2/3)

**x = ln (2/3)**