Question

Să se rezolve ecuația:
log₃x + logₓ9 = 3

Answer 1

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

Answer 2

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