Why does $\frac{1}{6e^{2y}}=\frac{1}{2x-8}$ in this context?

Question

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This is the context:

I tried substituting $y=3e^{2x}+4$ into $6e^{2y}$but I wasn't able to go any further. Does anyone what exactly is being done in the last step?

Answer 1

If you actually want to use a substitution to check your answer, then if $y=f^{-1}(x)$ and $f(x)=3e^{2x}+4$ then $$y=\dfrac12\log_e\left(\dfrac{x-4}{3}\right)$$

which will give $$\frac{1}{6e^{2y}} = \frac{1}{6e^{\log_e\left(\frac{x-4}{3}\right)}}= \frac{1}{6\left(\frac{x-4}{3}\right)}= \frac{1}{2x-8}$$

but this is not strictly necessary here

Answer 2

From the second line, $x= 3e^{2y} + 4$ $\Rightarrow 3e^{2y} = x-4$ $\Rightarrow 6e^{2y} = 2x-8$.

Answer 3

$3e^{2y} + 4 = x \implies 3e^{2y} = x - 4 \implies 6e^{2y} = 2x - 8 $

Therefore : $\frac{1}{2x - 8 }=\frac{1}{6e^{2y}}$