What is the answer to ln(1-e^x)+ln(1+e^x)=x ?

The answer to ln(1-e^x)+ln(1+e^x)=x is x=ln((sqrt(5)-1)/2)

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ln(1-e^x)+ln(1+e^x)=x

ln (1 - e^{x}) + ln (1 + e^{x}) = x

x = ln (\frac{5 - 1}{2})

Decimal Notation

x = - 0.48121 \dots

Solution steps

ln (1 - e^{x}) + ln (1 + e^{x}) = x

Apply log rules

ln ((1 - e^{x}) (1 + e^{x})) = x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{l n ((1 - e^{x}) (1 + e^{x}))} = e^{x}

Simplify

e^{l n ((1 - e^{x}) (1 + e^{x}))} : (1 - e^{x}) (1 + e^{x})

(1 - e^{x}) (1 + e^{x}) = e^{x}

Solve

(1 - e^{x}) (1 + e^{x}) = e^{x} : x = ln (\frac{5 - 1}{2})

x = ln (\frac{5 - 1}{2})

Verify Solutions:

x = ln (\frac{5 - 1}{2})

True

The solution is

x = ln (\frac{5 - 1}{2})

lo g_{5} (x - 6) = - lo g_{5} (x - 2) + 1

lo g_{5} (x - 4) = - 3

lo g_{2} (x - 3) + lo g_{2} (x - 2) = 1

ln (t - 2) = ln (8) - ln (t)

lo g_{10} (4 x) = lo g_{10} (7)

What is the answer to ln(1-e^x)+ln(1+e^x)=x ?
The answer to ln(1-e^x)+ln(1+e^x)=x is x=ln((sqrt(5)-1)/2)