Question

log in baza x din5+2log in baza5x din 5+3login baza25x din5=0

Answer 1

Răspuns:

logₓ5+2log₅ₓ5+3log₂₅ₓ5=0

Conditiile    de existenta   a    logaritmilor

x>0 x=/=1

5x=/=1   x=/=1/5

25x=/=1     x=/=1/25

aplici   proprietatea

logₐx=1/logₓa

1/log₅x+2/log₅5x+3/log₅25x=0

1/log₅x+2/(log₅5+log₅x)+3/(log₅25+log₅x)=0

1/log₅x+2/(1+log₅x)+3/(log₅5²+log₅x)=0

log₅x=y

1/y+2/(1+y)+3/(2log₅5+y)=0

1/y+2/(1+y)+3/(2+y)=0

(1+y)*(2+y)+2y(2+y)+3y(2+y)=0

(2+2y+y+y²+4y+2y²+6y+3y²=0

6y²+13y+2=0

Δ=169-48=121

y1=(-13-√121)/12=(-13-11)/12=-24/12=-2

log₅x=-2   x=5⁽⁻²⁾=1/25

y2=(-13+√121)/12=(-13+11)/12= -2/12=-1/6

log₅x=-1/6

x=5⁽⁻¹/⁶

Explicație pas cu pas: