Confusion regarding solving inequality $\log_2(x-1) \geq 3$. Now i got answer to be $[9,\infty]$. But i don't see why we have to take intersection with domain at the end?
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Intersection with which domain at the end? The answer is $[9,\infty)$. – Sarvesh Ravichandran Iyer Oct 21 '16 at 08:23
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Intersection with domain of original inequality i.e domain of log – J. Deff Oct 21 '16 at 08:42
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The domain of the logarithm function is $(0,\infty)$ Hence, it is only required that $x \in (1,\infty)$ for the function to be well defined. We already got $[9,\infty)$, and $[9,\infty) \cap (1,\infty) = [9,\infty)$, so even after taking intersection we are left with the same thing – Sarvesh Ravichandran Iyer Oct 21 '16 at 08:44
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But why take intersection is my question – J. Deff Oct 21 '16 at 08:48
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1Otherwise the logartithm would not make sense? For example, $\log (-x^2 - 3)$ cannot be defined at all. – Sarvesh Ravichandran Iyer Oct 21 '16 at 08:54
2 Answers
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First of all, take the necessary condition
$$x-1 > 0$$
hence
$$x > 1$$ otherwise you cannot proceed.
Now, exponentiate in base $2$:
$$2^{\log_2(x-1)} \geq 2^3$$
$$x-1 \geq 8$$
$$x \geq 9$$
Hence the domain is given by $[9, +\infty)$
Enrico M.
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This is a question of notation really. This also seems like a homework problem because you said that you got the answer to be [9, $\infty$] yet you don't see why you chose to include ] at the end?
There is a question thread that covers this quite extensively.
