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simplify; Log3 9x^2-Log3 7xy= Log3 9x/7y???
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\[\log_{3} (9x^2)-\log_{3}(7xy)=\log_{3}(9x/7y)\] ???? is the expression?
no the term after the = is my answer
it looks good to me
agreed
really???? wow
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how did u get this answer?
i used the logarithm rule for subtraction
9x^2/7xy =9x/7y
logx-logy=log(x/y)
this is the subtraction rule
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logarithmic properties: \[\log(ab)=\log(a)+\log(b)\] and \[\log(x^n)=n \log(x)\]
where log is with base 3
just make sure the bases of the logs are the same, or else those rules won't work.
so was i wrong???
No! you were right, I don't know what uzma did.
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