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I did solve, I got four solutions, but the book says there are only 3.

I considered the cases $| x - 3 | = 1$ or $3x^2 -10x + 3 = 0$.

I got for $x\leq 0$: $~2 , 3 , \frac13$

I got for $x > 0$: $~4$

Am I wrong? Is $0^0 = 1$ or NOT?

Considering the fact that : $ 2^2 = 2 \cdot2\cdot 1 $

$2^1 = 2\cdot 1$

$2^0 = 1$

$0^0$ should be $1$ right?

AndreasT
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Ricky
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    It depends. Sometimes $0^0$ is left undefined, sometimes it's assigned to equal $1$ (but never anything else). In this case, your textbook decided to not define $0^0$. – user236182 Dec 12 '15 at 13:53
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    You mean: for $x-3\le 0$ you got $2,3,1/3$ and for $x-3>0$ you got $4$. – user236182 Dec 12 '15 at 13:58

2 Answers2

1

First of all notice that $0^0\ne 1$

$$|x-3|^{3x^2-10x+3}=1\Longleftrightarrow$$ $$\ln\left(|x-3|^{3x^2-10x+3}\right)=\ln(1)\Longleftrightarrow$$ $$\ln\left(|x-3|\right)\left(3x^2-10x+3\right)=0\Longleftrightarrow$$ $$\ln\left(|x-3|\right)\left(x-3\right)\left(3x-1\right)=0$$

Split $\ln\left(|x-3|\right)\left(x-3\right)\left(3x-1\right)$ into seperate parts with additional assumptions:

So we got:

  • $$\ln\left(|x-3|\right)=0\Longleftrightarrow$$ $$|x-3|=1\Longleftrightarrow$$ $$x-3=\pm 1\Longleftrightarrow$$ $$x=\pm 1+3$$
  • $$x-3=0\to\text{not a valid solution}$$
  • $$3x-1=0\Longleftrightarrow$$ $$3x=1\Longleftrightarrow$$ $$x=\frac{1}{3}$$

So at the end we got 3 solutions:

$$x_1=\frac{1}{3}$$ $$x_2=2$$ $$x_3=4$$

Jan Eerland
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You are wrong $0^0=indeterminate$ you will get only three solutions. mod $| |$ gets opened with $\pm$ if we put it as '-'. We will get expression...=$-1$ so we can express negative number as raised to something only with a complex number here $i^2$ but we want real solutions so $...expression=1$ and you get three solutions. Hope its clear

Archis Welankar
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