Answer
$y = \dfrac{C^{\frac{1}{3}}(2x+1)^{\frac{1}{6}}}{(x+4)^{\frac{1}{9}}}$
Work Step by Step
RECALL:
(1) $r \cdot \ln{M}=\ln{(M^r)}$
(2) $\ln{M} + \ln{N} = \ln{(MN)}$
(3) $\ln{M} - \ln{N} = \ln{\left(\dfrac{M}{N}\right)}$
(4) $\ln{M} = \ln{N} \longrightarrow M=N$
(5) $a^{\frac{m}{n}}=\sqrt[n]{a^m}$
Using rule (1) above, the given expression is equivalent to:
$\ln{y^3}=\ln{(2x+1)^\frac{1}{2}}-\ln{(x+4)^{\frac{1}{3}}}+\ln{C}$
Use rule (3) above to obtain:
$\ln{y^3}=\ln{\left(\dfrac{(2x+1)^{\frac{1}{2}}}{(x+4)^{\frac{1}{3}}}\right)}+\ln{C}$
Use rule (2) above to obtain:
$\ln{y^3}=\ln{\left(\dfrac{(2x+1)^{\frac{1}{2}}}{(x+4)^{\frac{1}{3}}} \cdot C\right)}
\\\ln{y^3}=\ln{\left(\dfrac{C(2x+1)^{\frac{1}{2}}}{(x+4)^{\frac{1}{3}}} \right)}$
Use rule (4) above to obtain:
$y^3=\dfrac{C(2x+1)^{\frac{1}{2}}}{(x+4)^{\frac{1}{3}}}$
Take the cube root of both sides to obtain:
$y = \sqrt[3]{\dfrac{C(2x+1)^{\frac{1}{2}}}{(x+4)^{\frac{1}{3}}}}$
Use rule (4) above to obtain:
$y = \left(\dfrac{C(2x+1)^{\frac{1}{2}}}{(x+4)^{\frac{1}{3}}}\right)^{\frac{1}{3}}$
Use the rule $\left(\dfrac{a}{b}\right)^m=\dfrac{a^{m}}{b^{m}}$ to obtain:
$y = \dfrac{\left[C(2x+1)^{\frac{1}{2}}\right]^{\frac{1}{3}}}{\left[(x+4)^{\frac{1}{3}}\right]^{\frac{1}{3}}}$
Use the rule $(ab)^m = a^mb^m$ to obtain:
$y = \dfrac{C^{\frac{1}{3}}\left((2x+1)^{\frac{1}{2}}\right)^{\frac{1}{3}}}{\left((x+4)^{\frac{1}{3}}\right)^{\frac{1}{3}}}$
Use the rule $(a^m)^n=a^{mn}$ to obtain:
$y = \dfrac{C^{\frac{1}{3}}(2x+1)^{\frac{1}{6}}}{(x+4)^{\frac{1}{9}}}$
