Solve 3^x = x^9 for real x.

Step-by-step solution

1. 1

   Restrict the domain using the left side. $3^{x}>0$ for every real $x$, so $x^{9}$ must be positive too. Since $9$ is odd, $x^{9}>0$ forces

   $x>0$

   so only positive $x$ need be considered — and the logarithm is then legal.

2. 2

   Take logarithms to turn both sides into products. For $x>0$:

   $x\ln3=9\ln x\qquad\Longrightarrow\qquad\frac{\ln x}{x}=\frac{\ln3}{9}\approx0.122068$

   The problem has become: where does the curve $g(x)=\dfrac{\ln x}{x}$ hit a fixed horizontal level?

3. 3

   Spot the exact root. Try to make both sides a power of the same base. Since $27=3^{3}$,

   $27^{9}=\left(3^{3}\right)^{9}=3^{27}\ \checkmark$

   so $x=27$ solves the equation exactly — verified in exact integer arithmetic ✓. (By contrast $x=9$ fails: $3^{9}=19683$ but $9^{9}=387\,420\,489$, and $x=3$ fails too.)

4. 4

   Count the roots from the shape of $g$. Differentiating, $g'(x)=\dfrac{1-\ln x}{x^{2}}$, which is positive for $x<e$ and negative for $x>e$. So $g$ rises to a single maximum $g(e)=1/e\approx0.3679$ and then decreases to $0$. Since the target level $0.122068$ is below that maximum and above $0$, the horizontal line cuts the curve exactly twice — once on each side of $x=e$. The known root $27$ lies to the right, so a second root must lie in $(1,e)$.

5. 5

   Locate the second root numerically. Bisecting $g(x)=\ln3/9$ on $(1,e)$ converges to

   $x\approx1.1508248$

   Check: $3^{1.1508248}=3.5406499$ and $1.1508248^{9}=3.5406499$ ✓, agreeing to $10^{-13}$. So the equation has two real solutions — reporting only $x=27$ misses half the answer:

   $x=27\qquad\text{and}\qquad x\approx1.15082$