In [1]:
z1(x) = 5^x
In [2]:
z2(x) = x^5
In [3]:
S = solve(z1(x) == z2(x), x, to_poly_solve=True)
In [4]:
F(x, y, z) = x^2 - y^2 - z^2
In [5]:
implicit_plot3d(F(x, y, z) == 1,
    (x, -4, 4), (y, -4, 4), (z, -4, 4), viewer='tachyon')
Out[5]:
In [6]:
latex(S)
Out[6]:
\left[x = -\frac{5 \, \operatorname{W}({-\frac{1}{20} \, \sqrt{5} \log\left(5\right) - \frac{1}{20} i \, \sqrt{2 \, \sqrt{5} + 10} \log\left(5\right) + \frac{1}{20} \, \log\left(5\right)})}{\log\left(5\right)}, x = -\frac{5 \, \operatorname{W}({\frac{1}{20} \, \sqrt{5} \log\left(5\right) - \frac{1}{20} i \, \sqrt{-2 \, \sqrt{5} + 10} \log\left(5\right) + \frac{1}{20} \, \log\left(5\right)})}{\log\left(5\right)}, x = -\frac{5 \, \operatorname{W}({\frac{1}{20} \, \sqrt{5} \log\left(5\right) + \frac{1}{20} i \, \sqrt{-2 \, \sqrt{5} + 10} \log\left(5\right) + \frac{1}{20} \, \log\left(5\right)})}{\log\left(5\right)}, x = -\frac{5 \, \operatorname{W}({-\frac{1}{20} \, \sqrt{5} \log\left(5\right) + \frac{1}{20} i \, \sqrt{2 \, \sqrt{5} + 10} \log\left(5\right) + \frac{1}{20} \, \log\left(5\right)})}{\log\left(5\right)}, x = -\frac{5 \, \operatorname{W}({-\frac{1}{5} \, \log\left(5\right)})}{\log\left(5\right)}\right]

$[x = -\frac{5 \, \operatorname{W}({-\frac{1}{20} \, \sqrt{5} \log\left(5\right) - \frac{1}{20} i \, \sqrt{2 \, \sqrt{5} + 10} \log\left(5\right) + \frac{1}{20} \, \log\left(5\right)})}{\log\left(5\right)},$
$x = -\frac{5 \, \operatorname{W}({\frac{1}{20} \, \sqrt{5} \log\left(5\right) - \frac{1}{20} i \, \sqrt{-2 \, \sqrt{5} + 10} \log\left(5\right) + \frac{1}{20} \, \log\left(5\right)})}{\log\left(5\right)},$
$x = -\frac{5 \, \operatorname{W}({\frac{1}{20} \, \sqrt{5} \log\left(5\right) + \frac{1}{20} i \, \sqrt{-2 \, \sqrt{5} + 10} \log\left(5\right) + \frac{1}{20} \, \log\left(5\right)})}{\log\left(5\right)},$
$x = -\frac{5 \, \operatorname{W}({-\frac{1}{20} \, \sqrt{5} \log\left(5\right) + \frac{1}{20} i \, \sqrt{2 \, \sqrt{5} + 10} \log\left(5\right) + \frac{1}{20} \, \log\left(5\right)})}{\log\left(5\right)},$
$x = -\frac{5 \, \operatorname{W}({-\frac{1}{5} \, \log\left(5\right)})}{\log\left(5\right)}]$

In [ ]: