Question

Determine a solução das equações:
A) 2^x+3 + 2^x+1=40 B) 3^2x-1 = 3^5x+1 C) 2^2x+1=32

Answer 1

a) $2^x^+^3+2^x^+^1 = 40\\\\ 2^x*2^3+2^x*2^1=40\\\\ \boxed{2^x=y}\\\\ 8y+2y=40\\\\ 10y=40\\\\ y=\frac{40}{10}\\\\ \boxed{y=4}\\\\\ 2^x=y\\\\ 2^x = 4\\\\ 2^x=2^2\\\\ \boxed{x=2}$

b) $3^2^x^-^1 = 3^5^x^+^1\\\\ 2x-1 = 5x+1\\\\ 2x-5x = 1+1\\\\ -3x=2\\\\ \boxed{x=-\frac{2}{3}}$

c) $2^2^x^+^1 = 32\\\\ 2^2^x^+^1 = 2^5\\\\ 2x+1=5\\\\ 2x=5-1\\\\ 2x=4\\\\ x=\frac{4}{2}\\\\ \boxed{x=2}$

Answer 2

 
 
 
 

A) 2^x+3 + 2^x+1=40
    2³.2^x + 2.2^x=40
    8.2^x + 2.2^x=40
    10.2^x = 40
         2^x = 4
         2^x = 2^2
            x = 2


B) 3^2x-1 = 3^5x+1
       2x - 1 =  5x + 1
         - 1 - 1 = 5x - 2x 
            - 2 = 3x 
               x = - 2
                       3

C) 2^2x+1=32
     2^2x+1= 2^5
      2x + 1 = 5
         2x = 5 - 1
         2x = 4
           x = 2