Question

astept raspuns ....multumesc

Answer 1

- am atașat rezolvarea.

Answer 2

$E = 100^{\displaystyle \frac{1}{2}\lg 9-\lg 2} =100^{\displaystyle \lg 9^{\frac{1}{2}}-\lg 2} =100^{\displaystyle \lg \sqrt{9}-\lg 2} =$

$=100^{\displaystyle \lg 3-\lg 2} = 100^{\displaystyle \lg\frac{3}{2}} = (10^2)^{\displaystyle \lg\frac{3}{2}} = 10^{\displaystyle 2\lg\frac{3}{2}} =$

$=10^{\displaystyle \lg \left(\frac{3}{2}\right)^2} = 10^{\displaystyle \lg \frac{3^2}{2^2}} = 10^{\displaystyle \lg\frac{9}{4}} =10^{\displaystyle \log_{10}\frac{9}{4}} =\\ \\ = \boxed{\dfrac{9}{4}}$