While solving a problem, I reach to the conclusion that \( \ln(a-b) = \ln(b-a)\), which is actually false. Please tell me where I am wrong.

\[\large \ln(a-b) = \dfrac{1}{2}\ln(a-b)^2 = \dfrac{1}{2}\ln(b-a)^2 = \ln(b-a)\]

While solving a problem, I reach to the conclusion that \( \ln(a-b) = \ln(b-a)\), which is actually false. Please tell me where I am wrong.

\[\large \ln(a-b) = \dfrac{1}{2}\ln(a-b)^2 = \dfrac{1}{2}\ln(b-a)^2 = \ln(b-a)\]

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TopNewest\(\huge\frac{1}{2}\ln{(b-a)}^{2}=\ln\sqrt{{(b-a)}^{2}}=\ln\left| b-a \right| \) – Rohit Ner · 1 year, 8 months ago

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Domain of ln x is x>0.

First mistake: Taking ln of (a-b) \(\Rightarrow\) a-b>0, which simply means that ln of (b-a) is not defined\(\because b-a<0\)

Second mistake: ln \((x^{2n})\)=2n ln|x| \((x\epsilon Z)\) – Rishabh Cool · 1 year, 8 months ago

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Remember that \(\sqrt{x^2} = |x|\) and not \(x\).Its a kind of common misconception . – Nihar Mahajan · 1 year, 8 months ago

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