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Find the domain of the definition of the function

$$f(x)=\log_{0.3}\left(\log_{0.5}\left(\log_{0.8}\left(x^2-x+1\right)\right)\right)$$

My Try:

I assumed

$$f_1(x)=x^2-x+1$$ $$f_2(x)=\log_{0.8}(f_1(x))$$

$$f_3(x)=\log_{0.5}(f_2(x))$$

$$f(x)=\log_{0.3}(f_3(x))$$

Clearly $f_1(x)$ is Positive with Min value $0.75$. But for $f_3(x)$ to be defined $f_2(x) \gt 0$ which means $0 \lt f_1(x) \lt 1$. Hence

$$x^2-x+1 \lt 1$$ $\implies$ $$x \in (0 \:\: 1) \tag{1}$$

But for $f(x)$ to be defined $f_3(x) \gt 0$ which means $0 \lt f_2(x) \lt 1$

hence $$\log_{0.8}(f_1(x)) \lt 1$$ $\implies$

$$f_1(x) \lt 0.8$$ $\implies$

$$x^2-x+0.2 \lt 0$$ So

$$x \in (0.276 \:\: 0.723) \tag{2}$$

Taking intersection of $(1)$ and $(2)$ we have

$$Dom(f) = (0.276 \:\: 0.723)$$

Please let me know whether i am missing something

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1 Answer 1

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One of the inequalities is switched. In order for $\log_{0.8}f_1(x)<1$ to happen, we need $f_1(x)>0.8$, so it's $\text{dom}(f)=(0,0.276)\cup(0.723,1)$.

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  • $\begingroup$ No $f_1(x) \lt 0.8$ since $log_{10}x$ is an increasing function $\endgroup$
    – Ekaveera Gouribhatla
    Commented Dec 31, 2015 at 7:41
  • $\begingroup$ $log_{0.8}$ is decreasing $\endgroup$
    – JDZ
    Commented Jan 2, 2016 at 18:58
  • $\begingroup$ Oh yes you are absolutely correct, now i got my Err $\endgroup$
    – Ekaveera Gouribhatla
    Commented Jan 4, 2016 at 5:10

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