Question 1075860
solve the equation 3^x=2+3^-x.
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3^x = 2 + 3^-x
3^x - 2 - 3^-x = 0
Mutliply thru by 3^x
3^(2x) - 2*3^x - 1 = 0
Sub y for 3^x
y^2 - 2y - 1 = 0
*[invoke solve_quadratic_equation 1,-2,-1]
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Ignore the negative solution
3^x = 1 + sqrt(2)
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x*log(3) = log(1 + sqrt(2))
x = log(1 + sqrt(2))/log(3)
x =~ 0.80226