Someone please help fast, I am really lost with the law of logarithms and need help with these questions in picture

We'll use the following log rules

Part 1

Work Shown:

Ln(100) = Ln(4*25)

Ln(100) = Ln(2*2*5*5)

Ln(100) = Ln(2)+Ln(2)+Ln(5)+Ln(5) ... using log rule 1

Ln(100) = a+a+c+c

Ln(100) = 2a + 2c

Part 2

Work Shown:

Ln(2.4) = Ln(24/10)

Ln(2.4) = Ln[ (2*2*2*3)/(2*5) ]

Ln(2.4) = Ln[ (2*2*3)/5 ]

Ln(2.4) = Ln(2*2*3) - Ln(5) ... using log rule 2

Ln(2.4) = Ln(2)+Ln(2)+Ln(3) - Ln(5) using log rule 1

Ln(2.4) = a+a+b-c

Ln(2.4) = 2a+b-c

Part 3

Work Shown:

Ln(0.003) = Ln(3/1000)

Ln(0.003) = Ln[ 3/( 10^3) ]

Ln(0.003) = Ln(3) - Ln(10^3) ... log rule 2

Ln(0.003) = Ln(3) - 3*Ln(10) ... log rule 3

Ln(0.003) = Ln(3) - 3*Ln(2*5)

Ln(0.003) = Ln(3) - 3*( Ln(2) + Ln(5) ) ... log rule 1

Ln(0.003) = b - 3*( a + c )

Ln(0.003) = b - 3a - 3c

Ln(0.003) = -3a + b - 3c

Part 4

Work Shown:

All logs shown for this part are in base 2

log(5400) = log(54*100)

log(5400) = log(6*9*4*25)

log(5400) = log(2*3*3*3*2*2*5*5)

log(5400) = log(2^3*3^3*5^2)

log(5400) = log(2^3)+log(3^3)+log(5^2) ... log rule 1

log(5400) = 3*log(2)+3*log(3)+2*log(5) ... log rule 3

log(5400) = 3a + 3b + 2c

b

step-by-step explanation: