Show that $x=2\ln(3x-2)$ can be written as $x=\frac{1}{3}(e^{x/2}+2)$ Show that $x=2\ln(3x-2)$ can be written as $x=\dfrac{1}{3}(e^{x/2}+2)$.


Is there a rule for this?
Solve for the "other" $x$. Notice that:
$$ \begin{align*}
x &amp;= 2\ln(3x-2) \\
\dfrac{x}{2} &amp;= \ln(3x-2) \\
e^{x/2} &amp;= 3x-2 \\
e^{x/2}+2 &amp;= 3x \\
\dfrac{1}{3}(e^{x/2}+2) &amp;= x \\
\end {align*} $$
as desired.
exponential function is injective, so $x = y$ iff $e^x = e^y$. use this fact and then some elementary symbol manipulations should give you the desired result (if it's true of course)