Question 813798
4 over 3 1n m - 2 over 3 1n 8n - 1n m^3n^2 
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^ is commonly used for exponents.
It's LN, for natural log, no 1n.
There's nothing to solve, only combine and simplify.
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{{{(4/3)ln(m) - (2/3)ln(8n) - ln(m^3n^2)}}}
Many opportunities for error, wish us luck.
= {{{(1/3)ln(m^4) - (1/3)ln(64n^2) - ln(m^3n^2)}}}
= {{{(1/3)ln(m^4/(64n^2)) - ln(m^3n^2)}}}
= {{{(1/3)ln(m^4/(64n^2)) - (1/3)ln(m^9n^6)}}}
= {{{(1/3)ln(m^4/(64m^9n^8))}}}
= {{{(1/3)ln(1/(64m^5n^8))}}}
= {{{ln(root(3,(1/(64m^5n^8))))}}}
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= {{{ln(root(3,mn)/(4m^2n^3)))}}}