What is the answer to g= 1/((2+e^x)) ?

The answer to g= 1/((2+e^x)) is x=ln(1/g-2) {0<g< 1/2 }

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

g = \frac{1}{( 2 + e ^{x} )}

x = ln (\frac{1}{g} - 2) {0 < g < \frac{1}{2}}

Solution steps

g = \frac{1}{( 2 + e ^{x} )}

Multiply both sides by

2 + e^{x}

g (2 + e^{x}) = \frac{1}{2 + e ^{x}} (2 + e^{x})

Simplify

g (2 + e^{x}) = 1

Divide both sides by

g

2 + e^{x} = \frac{1}{g}

Move

2

to the right side

e^{x} = \frac{1}{g} - 2

Apply exponent rules

x = ln (\frac{1}{g} - 2)

Verify Solutions:

x = ln (\frac{1}{g} - 2) {0 < g < \frac{1}{2}}

The solution is

x = ln (\frac{1}{g} - 2) {0 < g < \frac{1}{2}}

(2^{x})^{2} - 3 \cdot 2^{x} + 2 = 0

x, 5^{x + 1} + 5^{1 - x} = 2 t

2^{x} = x^{3}

2^{1 - x} = \frac{1}{81}

3^{n} = 9 n

What is the answer to g= 1/((2+e^x)) ?
The answer to g= 1/((2+e^x)) is x=ln(1/g-2) {0<g< 1/2 }