Question

equação exponencial 3^x+2-3^x+1+3^x+3^x-1+3^x-3=16119

Answer 1

Deve ser:

3^(x+2) -3^(x+1) + 3^x +3^(x-1) +3^(x-3)=16119 não é?

3^(x).3^(2) -3^(x).3^(1) + 3^x +3^(x).3^(-1) +3^(x).3^(-3)=16119

3^(x)[3^(2) -3^(1) +1 +3^(-1) +3^(-3)]=16119

3^(x)[9 -3+1 +1/3 +1/27]=16119

3^(x)[7+1/3 +1/27]=16119

3^(x)[189+9 +1]=16119.27

3^(x)[199]=16119.27

3^(x)=16119.27/199

3^(x)=81.27

3^(x)=3^4.3^3

3^(x)=3^7

x=7