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

The answer to ln(e^x-1)=ln(e^{-x}-1) is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

Apply log rules

e^{x} - 1 = e^{- x} - 1

Solve

e^{x} - 1 = e^{- x} - 1 : x = 0

x = 0

Verify Solutions:

x = 0

False

The solution is

No Solution for x \in R

ln (2 x) = e^{2}

lo g_{4} (10) + lo g_{4} (x + 9) = 4

0.56 = lo g_{10} (x)

ln (x) + ln (2 x) = 3

lo g_{7} (x + 6) - 1 = - lo g_{7} (x)

What is the answer to ln(e^x-1)=ln(e^{-x}-1) ?
The answer to ln(e^x-1)=ln(e^{-x}-1) is No Solution for x\in\mathbb{R}