Question

sa se rezolve in multimea numerelor reale ecuatia (3^2x+1)-10*(3^x+1)+27=0

Answer 1

Explicație pas cu pas:

(3^2x+1)-10*(3^x+1)+27=0 ,notam 3^x=t>0

(t²+1)-10(t+1)+27=0

t²+1-10t-10-27=0

t²-10t-36=0

Δ=100+64=164⇒√164=2√41

t1=(10+2√41)/2=5+√41>0

t2=(10-2√41)/2=5-√41<0

deci 3^x=5+√41⇒x=log(3) din (5+√41)

Bafta!