Question 974196
{{{3^x + 1/3^x = 12}}}
Multiply by 3^x
{{{3^(2x) + 1 = 12*3^x}}}
{{{3^(2x) - 12*3^x + 1 = 0}}}
Sub u for 3^x
{{{u^2 - 12u + 1 = 0}}}
*[invoke solve_quadratic_equation 1,-12,1]
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3^x = 11.919
x*log(3) = log(11.919)
x = log(11.919)/log(3)
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Do the other 3^x root the same way.