Question

Vă rog frumos d) e) și f) de la exercițiul 11 ÎMI TREBUIE ACUM!!! DAU COROANĂ! ! !

Answer 1

d)  (2/5)^x : (2/5)^-3 = (2/5)^7
      x - (-3) = 7 
     x = 7 + (-3)
     x = 4 

e)  = [(2/3)^3]^x : (2/3)^4 = (2/3)^8
        3x - 4 = 8 
         3x = 8 + 4
         3x = 12 
         x = 12: 3 
         x = 4 

f)  15 - (2x-1) = 6 
     2x - 1 = 15 - 6 
     2x - 1 = 9
    2x = 9+1
    x  =10 : 2 
    x = 5 

Answer 2

   
[tex]\displaystyle\\ 11d)\\ \left(\frac{2}{5}\right)^x: \left(\frac{2}{5}\right)^{-3}=\left(\frac{2}{5}\right)^7\\\\ \left(\frac{2}{5}\right)^{x-(-3)}=\left(\frac{2}{5}\right)^7\\\\ \left(\frac{2}{5}\right)^{x+3}=\left(\frac{2}{5}\right)^7\\\\ x+3 =7\\\\ x = 7-3\\\\ x=\boxed{4}[/tex]


[tex]\displaystyle\\ 11e)\\ \left(\frac{8}{27}\right)^x:\left(\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^8\\\\ \left(\left(\frac{2}{3}\right)^3\right)^x:\left(\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^8\\\\ \left(\frac{2}{3}\right)^{3x}:\left(\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^8\\\\ \left(\frac{2}{3}\right)^{3x-4}=\left(\frac{2}{3}\right)^8\\\\ 3x-4=8\\\\ 3x=8+4\\\\ 3x=12\\\\ x=\frac{12}{3}=\boxed{4} [/tex]


[tex]\displaystyle\\ 11f)\\ \left(\frac{3}{5}\right)^{15}: \left(\frac{3}{5}\right)^{2x-1}=\left(\frac{3}{5}\right)^6\\\\ \left(\frac{3}{5}\right)^{(15-(2x-1))} =\left(\frac{3}{5}\right)^6\\\\ \left(\frac{3}{5}\right)^{(15-2x+1)} =\left(\frac{3}{5}\right)^6\\\\ \left(\frac{3}{5}\right)^{16-2x} =\left(\frac{3}{5}\right)^6\\\\ 16-2x=6\\\\ -2x = 6-16\\\\ -2x = -10~~~\Big| \times (-1) \\\\ 2x = 10\\\\ x= \frac{10}{2} = \boxed{5}[/tex]