Question Number 204273 by mustafazaheen last updated on 10/Feb/24
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{4}}\right)} } \\ $$$$\mathrm{Domain}\:\mathrm{f}\left(\mathrm{x}\right)\:=? \\ $$
Answered by Mathspace last updated on 11/Feb/24
![f(x)=(1/((x−1)^(ln((1/2))) ))=(1/((x−1)^(−ln2) )) =(x−1)^(ln2) =e^(ln2ln(x−1)) ⇒ D_f =]1,+∞[ en general la fonction x→x^a est definie pour x>0 (sur R)](https://www.tinkutara.com/question/Q204292.png)
$${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{−{ln}\mathrm{2}} } \\ $$$$=\left({x}−\mathrm{1}\right)^{{ln}\mathrm{2}} ={e}^{{ln}\mathrm{2}{ln}\left({x}−\mathrm{1}\right)} \:\Rightarrow \\ $$$$\left.{D}_{{f}} =\right]\mathrm{1},+\infty\left[\:\:{en}\:{general}\right. \\ $$$${la}\:{fonction}\:{x}\rightarrow{x}^{{a}} \:\:\:{est}\:{definie} \\ $$$${pour}\:{x}>\mathrm{0}\:\:\left({sur}\:{R}\right) \\ $$