with the steps please
expression = ln[⅓.(3x²-9x)] = ln[x(x-3)]
Is that enough simplification?
ln[3x*(x-3)*(1/3)]=ln[x(x-3)]=ln(x)+ln(x-3)
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expression = ln[⅓.(3x²-9x)] = ln[x(x-3)]
Is that enough simplification?
ln[3x*(x-3)*(1/3)]=ln[x(x-3)]=ln(x)+ln(x-3)