Question

3^x + 3^-x=10/3 cum se rezolva?

Answer 1

3^x + 3^(-x)=10/3 |·3=>3^x·3+ 3^(-x)·3=10/3 ·3 |·3^x=>
=>3^x·3·3^x·3+ 3^(-x)·3·3^x=10/3 ·3·3^x=>3^x·3^x·3+3·3^(-x+x)=10·3^x=>
=>3^x·3^x·3+3·3^0=10·3^x=>3^x·3^x·3+3·1=10·3^x=>3^x·3^x·3+3=10·3^x
Notam 3^x=y=>3^x·3^x·3+3=10·3^x=>y·y·3+3=10·y=>3·y^2-10·y+3=0->este o ecuatie de grad II
Aplicam Δ=b^2-4ac, unde b=-10, a=3 si c=3=>Δ=(-10)^2-4·3·3=>
=>Δ=100-36=>Δ=64=>ecuatia are solutii reale si distincte:
y₁=(-b+√Δ)/2·a=>y₁=[-(-10)+√64)/2·3=>y₁=(10+8)/6=>y₁=18/6=>y₁=3
y₂=(-b-√Δ)/2·a=>y₂=[-(-10)-√64)/2·3=>y₂=(10-8)/6=>y₂=2/6=>y₂=1/3=3^(-1)

Pentru y=3=>3^x=y=>3^x=3=>x=1
Pentru y=3^(-1)=>3^x=y=>3^x=3^(-1)=>x=-1

Deci x∈ {-1; 1}