Solve the following system of equations:
$$\ln(x)=4\ln(y)$$
$$\log_3x=2+2\log_3y$$
Here is my attempt to solve this problem. I rewrote the system as:
$e^4+y=x$ and $9=\frac{x}{y^2}$. When I tried to solve this system, I am not getting the correct answer, which is $(81, 3)$.
So, any help is greatly appreciated. Thank you!
Best Answer
You are not doing the conversion correctly.
$\ln x = 4\ln y \implies e^{\ln x} = e^{4 \ln y}$ and as $4\ln y = \ln y^4$ or because $e^{4\ln y} = (e^{\ln y})^4$ we have $e^{4\ln y} = y^4$ so $\implies x = y^4$.
$\log_3 x = 2+ 2\log_3 y\implies 3^{\log 3 x} =3^{2+2\log_3 y}\implies 3^{\log_3 x} = 3^2\cdot 3^{2\log_3 y}\implies x= 9\cdot 3^{2\log_3 y}$ and again because $2\log_3 y = \log_3 y^2$ or because $3^{2\log_3 y} = (3^{\log_3 y})^2$ we have $x = 9y^2$.
And that's that.
$x = y^4$ and $x = 9y^2$ so....
You need more practice (practice, practice, practice) on your log identities.
Practice practice practice.
Practice until your eyes bleed.