Relations and functions

Question

$$ g(x) = e^x-1, x\in\mathbb R\\ h(x) = \ln(\ln x),x>1\\ $$ By restricting the domain of $g$ to $(\alpha, \infty)$, where $\alpha \in \mathbb R$, find the smallest value of $\alpha$ in exact form such that the composite function $h\circ g$ exists. Define $h \circ g$.

My question: I don't understand why the restricted range of the g(x) is (1,∞)

What do you not understand? – michaelrccurtis Sep 04 '15 at 15:05 — michaelrccurtis, Sep 04 '15 at 15:05

score 1 · Accepted Answer · answered Sep 04 '15 at 15:16

1

$h\circ g(x)=h(g(x))=\ln(\ln(e^x-1))$ will be well defined if $\ln(e^x-1)$ is well defined and positive.

For this we need $e^x-1>1$ or equivalently $x>\ln2$.

answered Sep 04 '15 at 15:16

drhab

151,093

Where do you get ln2 from? Is it from 1=e^a -1 -> e^a =2 ? – erwinleonardy Sep 04 '15 at 15:19
$e^x-1>1\iff e^x>2\iff x>\ln2$ – drhab Sep 04 '15 at 15:21

Relations and functions

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