Can't solve exponential equation using logs? I can't figure out why my method isn't working.
I know it is possible to solve this using a substitution but I don't know when to use the substitution. In general when are you supposed to substitute for, say, u?
Here is how I did it;
We have the following function:
$3^{x-1}+3^{-x+1}$ and the slope of a point is given:  $\psi =\frac{8}{3}\ln(3)$
Calculate the coordinates of the point in which the slope equals $\psi$
My approach (Which is wrong):
$$\ln(3)3^{x-1}-\ln(3)3^{-x+1}= \frac{8}{3}\ln(3)$$
$$3^{x-1} - 3^{-x+1} = \frac{8}{3}$$
$$\log_{3}(3^{x-1})-\log_{3}{3^{-x+1}}=\log_{3}(\frac{8}{3})$$
$$(x-1)\log_{3}(3)-(-x+1)\log_{3}(3) = \log_{3}(\frac{8}{3})$$
$$(x-1)(1)-(-x+1)(1) =\log_{3}(\frac{8}{3})$$
$$x-1+x-1 =\log_{3}(\frac{8}{3})$$
$$2x-2=\log_{3}(\frac{8}{3})$$
$$2x = \log_{3}(\frac{8}{3})+2$$
$$x = \frac{1}{2}\log_{3}(\frac{8}{3})+1$$
Now I have a hard time with problems like these in general so all help is appreciated.
-Bowser

Here after your first line of taking the derivatives, set $u = x-1$
$3^{x-1} - 3^{-x+1} = \frac{8}{3}$
$3^{u} - 3^{-u} = \frac{8}{3}$
Furthermore set $v = 3^u$
$v - \frac{1}{v} = \frac{8}{3}$
$\frac{v^2 - 1}{v} = \frac{8}{3}$
$v^2 - 1= \frac{8v}{3}$
$v^2 - \frac{8v}{3} - 1=0$
$\Delta = \sqrt{ \frac{64}{9} + 4 }$
$v_{1,2} = \frac{\frac{8}{3} +- \sqrt{ \frac{100}{9} }}{2}$
$v_{1,2} = \frac{\frac{8}{3} +-  \frac{10}{3} }{2}$
$v_{1} = \frac{18}{6} = 3$
$v_{2} = \frac{-2}{6} = -0.33333...$
Since $v = e^u$, it can not be negative, so only $v = 3$
Now put back $u = ln(v)$ and $x = 1+u$
