College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.5 - Properties of Logarithms - 6.5 Assess Your Understanding: 95

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: $y = \dfrac{C^{\frac{1}{3}}(2x+1)^{\frac{1}{6}}}{(x+4)^{\frac{1}{9}}}$
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