How do you condense $1/3[lnx^9 - lny^12]$ ?

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Answer 1 · 2016-04-21T17:57:35

$[ln(x^9/y^12)]^(1/3)$

Use the Property: $log_b (x/y)=log_bx-log_by and log_bx^n=nlog_bx$

$1/3 [ln(x^9/y^12)]$

$[ln(x^9/y^12)]^(1/3)$

How do you condense 1/3[lnx^9 - lny^12]1/3[lnx^9 - lny^12]?