We know that \(e^{i\pi} = -1 \). So, \(\ln(-1) = i\pi \). Well, \(\ln(1) = 0\). So, I guess we can write \[\ln(-1) + \ln(-1) = \ln(-1\cdot -1) = \ln(1) = 0\]. But \[\ln(-1) + \ln(-1) = i\pi + i\pi = 2i\pi \]

What's the falacy?

We know that \(e^{i\pi} = -1 \). So, \(\ln(-1) = i\pi \). Well, \(\ln(1) = 0\). So, I guess we can write \[\ln(-1) + \ln(-1) = \ln(-1\cdot -1) = \ln(1) = 0\]. But \[\ln(-1) + \ln(-1) = i\pi + i\pi = 2i\pi \]

What's the falacy?

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## Comments

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TopNewestthe identity you used is applicable for numbers belonging to the set (0,infinity) only – Vivek Kushal · 3 years ago

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May be because the properties of ln is applicable only for numbers within its domain.. – Kushagra Jaiswal · 3 years ago

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– Gabriel Laurentino · 3 years ago

I thought the same. Thanks!Log in to reply