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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

Cro-Magnon
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    Log of a difference is not the difference of the logs. So there is an error in going from the second displayed line to the third. Instead, let $u=3^{x-1}$. – André Nicolas Mar 18 '16 at 11:53
  • @AndréNicolas When do you substitute for u in general? On what basis do you decide to substitute is what I'm asking. – Cro-Magnon Mar 18 '16 at 11:57
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    There are no general rules. But after a while you will be able to recognize relatives of this problem. In this case we get $u-\frac{1}{u}=\frac{8}{3}$, which is equivalent to a quadratic in $u$. Solve,, and you will know $3^{x-1}$, and then you can find $x-1$. – André Nicolas Mar 18 '16 at 12:00
  • I see.. I will practice this a lot from now on. Thanks for your input. – Cro-Magnon Mar 18 '16 at 12:02

2 Answers2

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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 $x = 1+u$

crbah
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A mistake was made in going from $3^{x-1}-3^{-x+1}=\frac{8}{3}$ to $\log_3(3^{x-1})-\log_3(3^{-x+1})=\log_3\left(\frac{8}{3}\right)$.

It looks as if implicitly the "law" $\log(a-b)=\log a-\log b$ was used. But that law is not correct.

Instead, let $u=3^{x-1}$. Then our equation is $u-\frac{1}{u}=\frac{8}{3}$, which with some manipulation turns into a quadratic equation. Solve for $u$, and then find $x$.

André Nicolas
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