# Where's the flaw?

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.

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

Note by Akhil Bansal
2 years, 3 months ago

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$$\huge\frac{1}{2}\ln{(b-a)}^{2}=\ln\sqrt{{(b-a)}^{2}}=\ln\left| b-a \right|$$

- 2 years, 3 months ago

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| $$(x\epsilon Z)$$

- 2 years, 3 months ago

Remember that $$\sqrt{x^2} = |x|$$ and not $$x$$.Its a kind of common misconception .

- 2 years, 3 months ago