MHB How Can You Rewrite an Equation to Express y as a Function of x?

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The discussion focuses on rewriting the equation \(3\ln{y}=\dfrac{1}{2}\ln{(2x+1)}-\dfrac{1}{3}\ln{(x+4)}+\ln{C}\) to express y as a function of x. Participants suggest using logarithmic properties to simplify the equation, leading to \( \ln(y) = \ln\left(\frac{C(2x+1)^{1/6}}{(x+4)^{1/9}}\right) \). After applying exponentiation, y is isolated as \( y = \frac{C_2\sqrt[6]{2x+1}}{\sqrt[9]{x+4}} \). The discussion highlights the importance of collaboration in simplifying complex equations. Overall, the process demonstrates how logarithmic manipulation can clarify the expression for y.
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$\tiny{6.5.95 Kamehameha HS}$

Express y as a function of x. $\quad C>0$
$3\ln{y}=\dfrac{1}{2}\ln{(2x+1)}-\dfrac{1}{3}\ln{(x+4)}+\ln{C}$
rewirte as
$\ln{y^3}=\ln{(2x+1)^{(1/2)}}-\ln{(x+4)^{(1/3)}}+\ln{C}$
then e thru and isolate y
i think
looks like it will be ugly
 
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$y = \dfrac{C_2\sqrt[6]{2x+1}}{\sqrt[9]{x+4}}$
 
karush said:
$\tiny{6.5.95 Kamehameha HS}$

Express y as a function of x. $\quad C>0$
$3\ln{y}=\dfrac{1}{2}\ln{(2x+1)}-\dfrac{1}{3}\ln{(x+4)}+\ln{C}$
rewirte as
$\ln{y^3}=\ln{(2x+1)^{(1/2)}}-\ln{(x+4)^{(1/3)}}+\ln{C}$
I would have said, instead,
$ln(y)= \ln((2x+1)^{1/6})- \ln((x+4)^{1/9})+ ln(C)$
and then use the properties of the logarithm
$ln(y)= \ln\left(\frac{C(2x+1)^{1/6}}{(x+ 4)^{1/9}}\right)$

then e thru and isolate y
i think
looks like it will be ugly
Now, it doesn't look so ugly!
 
well that certainly helped a lot
pays to get more input
 
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