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I am working on a test study guide and I can't seem to get the correct answer for this system of equations: \begin{align*} \ln(x) &= 3\ln(y) \\ \ 3^x &= 27^y \end{align*}

I'm not really sure how to go about solving this equation, so any help/direction would be much appreciated. Thank you in advance.

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    $\begingroup$ Are you sure those are the correct equations you were given? $\endgroup$ – user142299 Apr 14 '14 at 23:13
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    $\begingroup$ Maybe the second equation was supposed to be $3^x=27^y$? $\endgroup$ – David Apr 14 '14 at 23:15
  • $\begingroup$ Yes I just checked my guide again and the it is in fact 27^y not 27y. This makes it a lot more easier than I thought it was, my mistake! $\endgroup$ – user143171 Apr 14 '14 at 23:16
  • $\begingroup$ @user143171 In that case, see my answer. $\endgroup$ – user122283 Apr 14 '14 at 23:18
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Notice

$$ \ln x = 2 \ln y \iff \ln x = \ln y^2 \iff x = y^2 $$ and $$ 3^x = 27 y \iff 3^x = 3^3 y \iff 3^{x-3} = y \iff 3^{y^2 - 3} = y \iff \ln 3 ( y^2 - 3) = y \iff y^2 - 3 - \frac{y}{\ln 3} = 0 \iff y^2 - \frac{1}{\ln 3} y - 3 = 0$$

This is a quadratic equation, which you can easily solve.

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    $\begingroup$ $\ln 3 ( y^2 - 3) = y$ is wrong, it should be $(\ln 3) ( y^2 - 3) =\ln y$. The equation is no longer quadratic and it doesn't look easy. $\endgroup$ – David Apr 14 '14 at 23:13
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    $\begingroup$ And you should have $3\ln y$ in the first line not $2\ln y$. $\endgroup$ – user142299 Apr 14 '14 at 23:13
  • $\begingroup$ Please see the comment posted by the OP. $\endgroup$ – user122283 Apr 14 '14 at 23:18
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The first equation says that $x,y>0$ and then it's equivalent to $x=e^{\ln x}=e^{3\ln y}=e^{\ln y^3}=y^3$.

The second equation is equivalent to $x=3y$ by taking a logarithm, so we have $$3y=y^3\Longleftrightarrow y(y-\sqrt3)(y+\sqrt3)=0$$ and the only solution is $y=\sqrt3$, which gives $x=3\sqrt3$.

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This will look something like the rest of the answers (since the principles underlying the calculations are simple), but hopefully this avoids the "rejection" issue.

Starting from

$$ \ 3^x \ = \ 27^y \ \ \Rightarrow \ \ 3^x \ = \ (3^3)^y \ = \ 3^{3y} \ \ $$

we use the "common base" method you learn early on to establish that $ \ x = 3y \ $ . Replacing this in the first equation produces

$$ \ln (3y) \ = \ 3 \ \ln (y) \ \ \Rightarrow \ \ \ln \ 3 \ + \ \ln \ y \ = \ 3 \ \ln \ y \ \ \Rightarrow \ \ 2 \ \ln \ y \ = \ \ln \ 3 $$

$$ \Rightarrow \ \ \ln \ y \ = \ \frac{1}{2} \ln \ 3 \ = \ \ln \ 3^{1/2} \ \ \Rightarrow \ \ y \ = \ 3^{1/2} \ \ \Rightarrow \ \ x \ = \ 3 \ \cdot \ 3^{1/2} \ \ . $$

Arranging to exponentiate at the end can only produce a positive value, and thereby dodges the "spurious solution" issue.

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