Find all real solutions of 6^x = x^18.

Step-by-step solution

1. 1

   Sort out the domain. For $x>0$ both sides are positive. For $x<0$ the right side $x^{18}$ is still positive because the exponent is even, and $6^x>0$ too, so negative $x$ are allowed. At $x=0$ we get $1=0$, false, so $x=0$ is excluded.

2. 2

   Reduce the positive case to a single monotone-then-decreasing function. Taking logarithms of $6^x=x^{18}$ for $x>0$:

   $x\ln 6=18\ln x\quad\Longrightarrow\quad\frac{\ln x}{x}=\frac{\ln 6}{18}\approx 0.0995422$

   The function $g(x)=\tfrac{\ln x}{x}$ has $g'(x)=\tfrac{1-\ln x}{x^2}$, so it rises on $(0,e]$ and falls on $[e,\infty)$ with maximum $g(e)=\tfrac1e\approx 0.3679$. Since $0<0.0995<\tfrac1e$, the horizontal line meets it exactly twice.

3. 3

   Spot the exact large root. Try $x=36$: because $36=6^2$,

   $36^{18}=\left(6^2\right)^{18}=6^{36}$

   so $x=36$ satisfies the equation exactly. Confirming with $g$: $\tfrac{\ln 36}{36}=\tfrac{3.583519}{36}=0.0995422$ $\checkmark$.

4. 4

   Find the small positive root numerically. It must lie in $(1,e)$ since $g(1)=0$ and $g$ increases to $g(e)$. Bisection on $g(x)-0.0995422$ gives

   $x\approx 1.1176815$

   Check: $6^{1.1176815}\approx 7.4084$ and $1.1176815^{18}\approx 7.4084$ $\checkmark$.

5. 5

   Show there is exactly one negative root. Put $x=-t$ with $t>0$; the equation becomes $6^{-t}=t^{18}$, i.e. $h(t)=6^{t}t^{18}=1$. Since

   $\frac{d}{dt}\ln h(t)=\ln 6+\frac{18}{t}>0$

   $h$ is strictly increasing from $0$ to $\infty$, so $h(t)=1$ has exactly one solution. Bisection gives $t\approx 0.9131150$, i.e. $x\approx-0.9131150$; there $6^{x}\approx 0.194741$ and $x^{18}\approx 0.194741$ $\checkmark$.

6. 6

   Collect all three roots. One negative, two positive, and no others by the monotonicity arguments above:

   $x\approx-0.91311,\qquad x\approx 1.11768,\qquad x=36$