equation $\log_{2x}216=x$ Solve the equation $\log_{2x}216=x$ where $x$ is real. 
My attempt,
By trial and error $x=3$ as $6^3=216$
Is there any systematic way to solve it?
 A: Noting that $$\log_{2x}(216) = \frac{\ln(216)}{\ln(2x)}$$ we can say \begin{align}\log_{2x}(216) &= x\\
\frac{\ln(216)}{\ln(2x)}&=x\\
\ln(216)&=x\ln(2x)\\
e^{\ln(216)}&=e^{x\ln(2x)}\\
e^{\ln(216)}&=e^{\ln((2x)^x)}\\
216&=(2x)^x
\end{align}
We can then note that $216=6^3$ and so \begin{align}6^3&=(2x)^x\\
(2\times 3)^3&=(2x)^x\end{align}
It is then trivial to see that $x=3$

We could also calculate this last step by noting that the exponent must be half the base.
We can then brute force this with integers:
Starting with an exponent of $1$, we have
\begin{align}2^1&=2\\
4^2&=16\\
6^3&=216\end{align}
And so our exponent, and thus $x$, must be $3$
A: The variable in the base is difficult to deal with. You can apply the change of base formula to find $$\frac{\log 216}{\log 2x} = x$$ where the logarithm is with respect to any base at all. However, since you already noted that $6^3 = 216$ it would be a good idea to use a logarithm base $6$. This leads to $$x = \frac{\log_6 216}{\log_6 2x} = \frac{3}{\log_6 2x}.$$ This simplifies to $$x \log_6 2x = 3.$$
Mixed expressions like this are not generally solvable in explicit terms, but since the functions $y = x$ and $y = \log_6 2x$ are both increasing, there is at most one solution. The first guess $2x = 6$ that makes $\log_6 2x = 1$ and $x = 3$ happily provides the solution.
A: These types of questions are almost impossible to solve in the "normal" method, i.e by rearranging the powers or some other similiar way. However in this particular case of 216,it is the quickest path to the answer.
What you can do is to take both sides to the exponent of $2x$ and thereby obtain
$(2x)^x=216$
you can easily see that $216= 6^x=(2 \times 3)^3$
Relating the above two, we can clearly see that $x=3$.
A: The equation is equivalent to $(2x)^x=216$ or, after squaring and setting $2x=t$,
$$
t^t=6^6
$$
More generally, let's consider $t^t=a$. This can be rewritten as
$$
t\log t-\log a=0
$$
(natural logarithm). Consider the function $f(t)=t\log t-\log a$. Then
$$
f'(t)=\log t+1
$$
which vanishes only for $t=1/e$, is positive over $(1/e,\infty)$ and negative over $(0,1/e)$. Thus $f$ has a minimum at $1/e$. Now
$$
f(1/e)=-\frac{1}{e}-\log a
$$
so the equation will have


*

*two solutions if $f(1/e)<0$

*one solution if $f(1/e)=0$

*no solution if $f(1/e)>0$
The first case happens when
$$
\frac{1}{e}+\log a>0
$$
that is $a>e^{1/e}\approx1.444667861$, which is certainly the case when $a=6^6$.
One solution is of course $t=6$, that is, $x=3$. The other solution $t'$ cannot be determined analytically, but from the above discussion we can conclude that $t'<1/e$, so $x'<1/(2e)$. In particular $x'$ cannot be integer.
A: No, not really. It certainly helps to note $216 = 6^3$; this reduces the possible solutions you have to check (e.g. $x=5$ is obviously not a solution), but in general these kind of equations are only solvable with the help of Lambert W functions.
