Question

oblicz

a) log₃(x²-5x+6)=0

b) log₃(x²-5x+6)=0

c) log₂(x²-3x)= 2

d) log₃ (x+1)= 2

Answer 1

a)

log3(x2 - 5x + 6) = log3(3^0)

x^2 - 5x + 6 = 1

x^2 - 5x + 5 = 0

delta = 25 - 20 = 5

x1 = (5 - pierwiastek z 5) / 2

x2 = (5 + pierwiastek z 5) / 2

 

b)

to samo równanie co w a)

 

c)

log2(x^2 - 3x) = log2(2^2)

x^2 - 3x = 4

x^2 - 3x - 4 = 0

delta = 9 + 16 = 25

x1 = (3 - 5) / 2 = -1

x2 = (3 + 5)/2 = 4

 

d)

log3(x + 1) = log3 (3^2)

x + 1 = 9

x = 9 - 1

x = 8

 

Prosze :)

Answer 2

log₃(x²-5x+6)=0

3^0=x²-5x+6

1=x²-5x+6

x²-5x+6-1=0

x²-5x+5=0

delta=25-20=5

x1-5-pierwiastek z 5/2

x2=5+pierwiastek z 5/2

 

log₃(x²-5x+6)=0

to jest to samo co punkt a

 

log₂(x²-3x)= 2

2^2=x²-3x

4=x²-3x

x²-3x-4=0

delta=9+16=25

x1=3-5/2=-2/2=-1

x2=3+5/2=8/2=4

 

log₃ (x+1)= 2

3^2=x+1

9=x+1

x=8