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log_{3}0.1+2log_{3}X= log_{3}2+log_{3}5 slove for x.
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log_{3}0.1+2log_{3}X= log_{3}2+log_{3}5 \[\rightarrow \log_{3}0.1+\log_{3}X ^{2}= \log_{3}10 \]
subtract from both sides log3(0,1)
on the right side will get log3(10/0,1)
and result x2 =1 so x1,2= +/- 1
ok... thanks
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