Question

Cat este x^{lgx} ? este logaritm in baza 10

Answer 1

$\text{Al k+1 termen dintr-o dezvoltare este :} \boxed{T_{k+1} =C_n^k \cdot a^{n-k} \cdot b^k}\\ \text{Din diferenta coeficientilor binomiali il putem afla pe n:}\\ C_n^2 -C_n^1 =27\\ \dfrac{n(n-1)}{2}-n =27\\ n^2-n-2n=54\\ n^2-3n-54=0\\ \Delta =9+4\cdot 54= 9+216 =225 \Rightarrow \sqrt{\Delta} =15\\ n_1= \dfrac{3+15}{2}=9\\ n_2=\dfrac{3-15}{2}= -6\notin\mathbb{N}\text{,deci solutia aceasta nu convine}\\ \text{Asadar}\ T_{k+1} =C_{9}^k\cdot \left(\dfrac{1}{\sqrt[8]{x}}\right)^{9-k}\cdot \left(x^{\lg x}\right)^k$

$T_2=C_9^1\cdot \left(\dfrac{1}{\sqrt[8]{x}}\right)^8 \cdot \left (x^{\lg x}\right)=9\cdot \dfrac{1}{x}\cdot x^{\lg x}= 9 \cdot x^{lg x-1} \\ T_2=900\\ 9\cdot x^{lg x-1} =900\\ x^{\lg x-1 }=100 |\lg()\\ (\lg x-1)\cdot \lg x=2\\ \text{Notam } \lg x=t\\ (t-1)\cdot t=2\\ t^2-t-2=0\\ (t-2)(t+1)=0\Rightarrow t\in \{2,-1\}\\ i)\text{Pentru }t=2\Rightarrow x=100$
$ii)\text{Pentru } t=-1 \Rightarrow x=\dfrac{1}{10}\\ S:x\in \left\{100,\dfrac{1}{10}\right \}$