Using the substitution $u=3^x$, or otherwise, solve, correct to 3 significant figures, the equation $3^x=2+3^{-x}$
My attempt:
$u=2+u^{-1}$
$u-u^{-1}=2$
$ {{u^2-1}\over {u}}=2$
$\log ({{u^2-1}\over {u}})=\log (2)$
$\log ({u^2-1}) - \log ({u}) =\log (2)$
$\log ({(u-1)(u+1)}) -\log (u) = \log (2)$
$\log {(u-1)} + \log {(u+1)} -\log (u)= \log (2)$
Then what can I do please ?
$\endgroup$ 43 Answers
$\begingroup$Once you get to $$\frac{u^2-1}{u}=2,$$ note that it can be solved as a quadratic equation: $$u^2-1=2u.$$ Solve it for $u$ as usual: $u=whatever$.
Finally, substitute back $u = 3^x$ and take logarithms: $3^x=whatever$, so that $x=\log_3(whatever)$.
$\endgroup$ 2 $\begingroup$Another way is to see the $\sinh$ function : $$3^x-3^{-x} =2 \sinh(x\ln(3)) = 2$$ So you get : $$x = \frac{\operatorname{arcsinh}(1)}{\ln(3)}= \frac{\ln(1+\sqrt{1+1^2})}{\ln(3)} = \frac{\ln(1+\sqrt{2})}{\ln(3)}$$
$\endgroup$ 1 $\begingroup$Hint: multiply everything by $3^x$, make the substitution and then it will became an equation of second degrees.
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