If you have x both in the exponent and the base of exponentiation, you should use Lambert W (productlog) function to get all real solutions. Quick answer is that there are two real solutions: x1 = e^(-productlog(-1, -ln(3)/9)) = 27 x2 = e^(-productlog(0, -ln(3)/9)) ≈ 1.151 because -ln(3)/9 is between -1/e and 0.

With hindsight, this seems a little over-complicated. As I should have remembered, it often helps to take both sides to a power, in this case 3. (3^x)^3 = (x^9)^3 3^(3x) = x^27 (3^3)^x = x^27 27^x = x^27 which makes it "obvious" that x=27. Except there is another solution.

As I follow, each step makes sense until we get the answer. I'm struggling to find any way to guess which approach to take?

How about finding the second root?

Stare at the given equation for a few seconds and see that x slightly greater than 1 is a solution. Also, since x is raised to an odd power it can't be negative because 3^x must be positive. Use fixed-point iteration to solve for x←(3^x)^(1/9). Starting at x=1 gives x=1.15082482... after a few iterations. For a large value of x start with x*ln3=9lnx and solve for x←9lnx/ln3. Starting at x=50 gives x=27.00000000... after a few iterations. This example shows that different forms of fixed-point iteration may be required to get different solutions.

(x+3x-3)

My solution is slightly different. Let x=3ᵃ Then 3^(3ᵃ)=(3ᵃ)⁹ 3^(3ᵃ)=3⁹ᵃ As 3>1, 3ᵃ=9a By inspection, a=3 is a solution, so x=3³=27 is a solution to the original equation, but there may be others. 3ᵃ=9a 1=9a3⁻ᵃ a3⁻ᵃ=1/9 a exp(-a ln3)=1/9 -a ln 3 exp(-a ln3)=-1/9 ln3 As -1/9 ln3-1/e there are two real solutions: -a ln3=W₀(-1/9 ln3) where -1

(x+3x-3)