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astept raspuns ....multumesc

Astept Raspuns Multumesc class=

Răspuns :

- am atașat rezolvarea.

Vezi imaginea Newton13

[tex]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} =[/tex]

[tex]=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}} =[/tex]

[tex]=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}}[/tex]