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I have easily solved a lot of log equations which do not involve the modulus function. I know the definition of the modulus function. But, how does this function affect the solutions of a equation? The given equation is an example.Can someone explain me the approach to solving equations having modulus by using this example (detailed explanation will be appreciated)?


The domain gives $x\geq0$.

Now for $0\leq x\leq1$ we get an identity.

For $x>1$ the $|.|$ disappears and we obtain: $$(2\sqrt{x}-1)^2=8\sqrt{x}-7,$$ which gives also $x=4$ and we get the answer: $$[0,1]\cup\left\{4\right\}$$

  • $\begingroup$ So, I have to check the domain of x before solving equations with modulus, right? $\endgroup$ – Mriganka Parasar May 16 '17 at 3:55
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    $\begingroup$ @Mriganka Parasar Yes, of course, but also we need to solve it. $\endgroup$ – Michael Rozenberg May 16 '17 at 3:57
  • $\begingroup$ Did you get the identity by observation? $\endgroup$ – Mriganka Parasar May 16 '17 at 4:11
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    $\begingroup$ @Mriganka Parasar Because $|x|=x$ for $x\geq0$ and $|x|=-x$ for $x\leq0$. In our case for $0\leq x\leq1$ we obtain: $\sqrt{x}+|\sqrt{x}-1|=\sqrt{x}-\sqrt{x}+1=1$. $\endgroup$ – Michael Rozenberg May 16 '17 at 4:12
  • $\begingroup$ Oh, thanks! Actually, you wrote 0 twice by mistake, so I was confused. Now, its fine. $\endgroup$ – Mriganka Parasar May 16 '17 at 4:14

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