MATHEMATICS PROBLEM SOLVER

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Saturday 25 April 2015

LOGARITHMS 18

If Log_b343 – log0.001 = 8; find b

Solution

Log_b343 – log0.001 = 6; [0.001=10^-3]

Log_b343 – log10^-3= 6

Log_b343 – (-3log10)= 6

Log_b343 + 3log10= 6 [remember log 10 = 1]

Log_b343 + (3 x 1) = 6

Log_b343 + 3 = 6

Log_b343 = 6 - 3

Log_b343 = 3 [Change it into exponential notation]

343 = b³

7³ = b³[ 243 = 7³ by prime factorization]

7³ = b³[ bases are same on both sides, so they cancel out ]

b = 7

Hence b = 7

TRY THIS..............

If Log_a64 – log0.00001 = 11; find a

MATHEMATICS PROBLEM SOLVER

Saturday 25 April 2015

LOGARITHMS 18

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