how would you solve for the tangent line to the graph of ???? = ????^X which has a slope of 3?

Slope of the tangent:
dy/dx = 3
eˣ = 3
x = ln(3)

When x = 3:
y = e¹ⁿ⁽³⁾
y = 3
Hence, the point of contact = (ln(3), 3)

Equation for the tangent line (point-slope form):
y - 3 = 3(x - ln(3))
y - 3 = 3x - 3ln(3)
3x - y + 3 - 3ln(3) = 0

Two parts: Find the point of tangency that has a slope (derivative) of 3:

    dy/dx = 3
    d(e^x) / dy = 3
    e^x = 3
    x = ln 3
    y = e^x = e^(ln 3) = 3

Now that you know the line passes through (ln 3, 3) with slope m=3, it's an Algebra 1 problem.  Point slope formula, in particular:

    y - 3 = 3(x - ln 3)

...and that's how I'd derive the equation that Ted S gave.

y - 3 = 3 ( x - ln 3)