How do you condense $\frac{1}{3} [{ln x}^{9} - {ln y}^{12}]$ $1/3[lnx^9 - lny^12]$ ?

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How do you condense $\frac{1}{3} [{ln x}^{9} - {ln y}^{12}]$ $1/3[lnx^9 - lny^12]$ ?

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Answer 1 · 2016-04-21T17:57:35

${[ln (\frac{x^{9}}{y^{12}})]}^{\frac{1}{3}}$ $[ln(x^9/y^12)]^(1/3)$

Explanation:

Use the Property: ${log}_{b} (\frac{x}{y}) = {log}_{b} x - {log}_{b} y and {log}_{b} x^{n} = n {log}_{b} x$ $log_b (x/y)=log_bx-log_by and log_bx^n=nlog_bx$

$\frac{1}{3} [ln (\frac{x^{9}}{y^{12}})]$ $1/3 [ln(x^9/y^12)]$

${[ln (\frac{x^{9}}{y^{12}})]}^{\frac{1}{3}}$ $[ln(x^9/y^12)]^(1/3)$

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How do you condense $\frac{1}{3} [{ln x}^{9} - {ln y}^{12}]$ $1/3[lnx^9 - lny^12]$ ?

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