Ho to Solve 2log3-log27+log9?

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Answer 1 · 2018-03-12T12:24:06

#log3#

#2log3-log27+log9#

Before we start do you notice something? #3#, #27#, and #9# are all powers of #3#.

#2log3-log27+log9#

Make the base #3#,

#2log3-log3^3+log3^2#

Using logarithm of a power law #(logx^n=nlogx)#,

#2log3-3log3+2log3#

Simplify,

#log3#