Hi,
mein Vorschlag:
2lg(x+3) = lg(x+1) + 1 |mit 1 = lg(10) und b*lg(a) = lg(a^b)
lg((x+3)^2) = lg(x+1) + lg(10) |mit lg(ab) = lg(a)+lg(b)
lg((x+3)^2) = lg(10x+10) |10 anwenden
(x+3)^2 = 10x+10
x^2+6x+9 = 10x+10 |-10x-10
x^2-4x-1 = 0 |pq-Formel
x1 = 2-√5 ≈ -0,236
x2 = 2+√5 ≈ 4,236
Alright?
Grüße