fie 5^ x³=5
atunci ecuatia devine
5t-5/t=24 a amplificam cu t
5t²-24t-5=0
t1,2= (24+-√(576+100))/10=(24+-26)/10
t1=-1/5
t2=5
5^(x1)³=-1/5 dar 5^f(x)>0, ptca este functie exponentiala, deci nu avem solutii pt x³
5^x2³=5 =5^1 ⇒ (x2)³=1 x=1
verificare
5^(1+1) -5^(1-1)=5²=5^0=25-1=24 adevarat
