Someone please help fast, I am really lost with the law of logarithms and need help with these questions in picture
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We'll use the following log rules
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Part 1
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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
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Part 2
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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
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Part 3
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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
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Part 4
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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