Question

Sa se determine valorile reale pozitive ale numaruli x, stiind ca lg√x , $\frac{3}{2}$ si lg x sunt tri termeni consecutivi ai unei progresi aritmetice

Answer 1

$\displaystyle \frac{3}{2} = \frac{lg \sqrt{x} +lg~x}{2} \Rightarrow 2 \cdot \frac{3}{2} =lg \sqrt{x} +lg~x \Rightarrow \frac{6}{2} =lg \sqrt{x} +lg~x \Rightarrow \\ \Rightarrow 3=lg \sqrt{x} +lg~x \Rightarrow 3= \frac{1}{2}lg~x+lg~x \Rightarrow 3= \frac{lg~x+2lg~x}{2} \Rightarrow \\ \Rightarrow 3= \frac{3lg~x}{2} \Rightarrow 3 \cdot 2=3lg~x \Rightarrow 6=3lg~x \Rightarrow lg~x= \frac{6}{3} \Rightarrow \\ \Rightarrow lg~x=2 \Rightarrow x=10^2 \Rightarrow x=100$