Implicit logarithmic differentiation to find the horizontal tangents of an exponential function

Question

The graph of $y = 6{(3{x}^2)}^x$ has two horizontal tangent lines. Find equations for both of them.

$$ \\ \begin{align} \\ y &= 6{(3{x}^2)}^x \\ y &= 6 \cdot {3}^x \cdot {x}^{2x} \\ \ln{y} &= \ln{(6 \cdot {3}^x \cdot {x}^{2x})} \\ \ln{y} &= \ln{6} + \ln{{3}^x} + \ln{{x}^{2x}} \\ \ln{y} &= \ln{6} + x\ln{3} + 2x\ln{{x}^{2x}} \\ \frac{d}{dx}[\ln{y}] &= \frac{d}{dx}[\ln{6} + x\ln{3} + 2x\ln{x}] \\ \frac{y'}{y} &= 0 + \ln{3} + 2\ln{x} + \frac{2x}{x} \\ y' &= y(2\ln{x} + \ln{3} + 2) \\ y' &= 6{(3{x}^2)}^x(2\ln{x} + \ln{3} + 2) \\ 0 &= 6{(3{x}^2)}^x(2\ln{x} + \ln{3} + 2) \\ \frac{0}{6{(3{x}^2)}^x} &= 2\ln{x} + \ln{3} + 2 \\ 0 &= 2\ln{x} + \ln{3} + 2 \\ -(\ln{3} + 2) &= 2\ln{x} \\ -\frac{\ln{3} + 2}{2} &= \ln{x} \\ e^{-\frac{\ln{3} + 2}{2}} &= x \\ \end{align} $$

I've made an error at some point, but I'm not certain what error I'm making.

The $x$ value I've found doesn't appear to satisfy the conditions of the prompt.

Insight?

Answer 1

Hint: $$ 2\ln x+\ln 3 +2 =0 \Rightarrow \ln x^2=-2-\ln 3 \Rightarrow x^2= \dfrac {1}{3e^2} \Rightarrow x=\pm \dfrac{1}{e \sqrt{3}} $$

Answer 2

You haven't made a mistake in the answer you found.

$$\exp\left({\frac{-\ln3 - 2}2}\right)= \exp \left({\frac {-\ln 3}2 - \frac 2 2 }\right)$$

$$ = \exp (-1) \exp \left({\frac {-\ln 3}2}\right)$$

$$ = e^{-1} \left({e^{\ln 3}}\right)^{-1/2}$$

$$ = e^{-1} 3^{-1/2}$$

$$ = \frac 1 {e \sqrt 3} $$

The other answer is missing because you simplified $\ln x^{2x} = 2x \ln x $ instead of the correct $2x \ln |x|$.

edit: unfortunately, these $x$-coordinates aren't the end of your answer, because your assignment asks you to find the equation of the tangent lines at those points!