Logaritmikus egyenletek

lgx = 3⋅lg72 + 2⋅lg54 − 4⋅lg108

x = 8

lgx = lg2 + lg8

x = 16

lgx = 3⋅lg2 + lg8

x = 64

lgx = 1 − lg2

x = 5

lg(10x − 2) − 2⋅lg(x + 1) = lg2

x₁ = 1;     x₂ = 2

lgx + lg3 = 2⋅lg4 − lg2

x = 8/3

lgx = 2⋅(lg3 + lg2)

x = 36

2 − lg₃x = lg₃2 + lg₃4 + lg₃5

x = 9/40

lgx = 3 − lg5

x = 200

2⋅lg(x+3) − lgx = lg(25x+3) − lg10

x₁ = 5;     x₂ = −1,2 Negatív számnak nincs logaritmusa, tehát ez nem megoldás!!!

lg(x + 1) + lg(x − 1) = lg8 + lg(x − 2)

x₁ = 3;     x₂ = 5

log₃√x − 2 − log₃(x − 5) + log₃2 = 0

x₁ = 3;     x₂ = 11

lgx − lg(x − 3) = 1

x = 10/3

lg(x² − 5x − 9) − lg(2x − 1) = 0

x₁ = 8;     x₂ = −1 Ez hamis gyök!!!

lg²x − lgx² = −1

x = 10