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Question
Express the following in terms of log 2 and log 3: log128
Sum
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Solution
log128
=log (3 x 22)8
= 8log (3 x 22)
= 8[log 3 + log22]
= 8[log 3 + 2log 2].
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Expansion of Expressions with the Help of Laws of Logarithm
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