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Să se rezolve ecuația:
log₃x + logₓ9 = 3


Răspuns :

x≠1   x>0l
log(3)x+log(x)9=3
log(3)x+log(x)3²=3
log(3)x=2log()3=3
substitutie
log(3)x=y  
ecuatia   devine
y+2/y=3
y²-3y+2=0
y1=1 =>   log(3)x=1  =>   x=3
y2=          log(3)x=2  x=3²=9

[tex]log_3x+log_x9=3[/tex]
Cond:
x>0 x∈(0,∞)
x≠1
x(0,∞)\{1}
[tex]log_3x+ \frac{log_39}{log_3x}=log_3x+ \frac{2}{log_3x} =t+ \frac{2}{t} =3\ \textless \ =\ \textgreater \ [/tex]
Notam  [tex]log_3x[/tex] cu t, t>0
t²-3t+2=0
Δ=9-8=1
t1=2 t2=1
[tex]log_3x[/tex]=2=>x=9
[tex]log_3x[/tex]=1 =>x=3
S={3,9}