Is that the base you want?

Using properties of logarithms, the equation can be rewritten as:

[log(base 3)(x)]^2 = (log(base 3)(3))*(log(base 3)(3^4)) = 4

Taking the positive and negative roots and then anti-logs yields x = 3^2 or x = 3^-2. Hence, the required solution is 1 + 1/9 = 10/9

$\begin{aligned} \log_x 3 & = \log_{81} x \\ \frac {\log 3}{\log x} & = \frac {\log x}{\log 81} = \frac {\log x}{4\log 3} \\ \implies \log^2 x & = 4 \log^2 3 \\ \log x & = \pm \log 9 \\ \implies x & = 9, \frac 19 \end{aligned}$

Therefore $9 \times \dfrac 19 + \dfrac 19 = \boxed{1.111...}$ .

let $log_x 3 = log_{81} x = a$

$log_x 3 = a$

$3=x^a$

$log_{81} x=a$

$81^a = x$

$3^{4a} = x$

Now, substitute

$(x^a)^{4a}=x$

$x^{4a^2}=x$

$4a^2 = 1$

$a^2=\dfrac{1}{4}$

$a=\pm \dfrac{1}{2}$

Now, solve for $x$

$x=81^{\frac{1}{2}}=9$

or

$x=81^{-\frac{1}{2}}=\dfrac{1}{81^{\frac{1}{2}}}=\dfrac{1}{9}$

The required answer is

$9 \times \dfrac{1}{9}+\dfrac{1}{9}\approx \boxed{1.111}$