Monday, 24 June 2013

LOGARITHMS - 2



If Logb125 – log 0.00001 = 8; find b

Solution

Logb125 – log 0.00001 = 8;            [0.00001=10-5 ]

Logb125 – log 10-5 = 8      

Logb125 – (-5log 10) = 8

Logb125 + 5log 10 = 8   [remember log 10 = 1]

Logb125 + (5 x 1) = 8

Logb125 + 5 = 8

Logb125 = 8 - 5

Logb125 = 3     [Change it into exponential notation]

125 = b3

53 = b3

b = 5


Hence b = 5       


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