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.

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

Easy Math Editor

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Comments

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

Rohit Ner - 5 years, 4 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| $(x\epsilon Z)$

Rishabh Jain - 5 years, 4 months ago

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

Nihar Mahajan - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$