Exponential Equation Example I have got the example given below from a friend, but I am not able to start solving it.
$$(x+24)^{1/x}=3$$
When $x=3$ does satisfy the equation. Any idea is welcome. 
 A: To solve
$$
(x+24)^{1/x}=3\tag1
$$
set $u=x+24$, then take logs:
$$
\begin{align}
\log(u)&=\log(3)(u-24)\tag2\\
\log(\log(3)u)&=\log(3)(u-24)+\log(\log(3))\tag3\\
\log(\log(3)u)-\log(3)u&=-24\log(3)+\log(\log(3))\tag4\\
-\log(3)u\,e^{-\log(3)u}&=-3^{-24}\log(3)\tag5\\
\end{align}
$$
Explanation:
$(2)$: take the log of $(1)$ and multiply by $u-24$
$(3)$: add $\log(\log(3))$
$(4)$: subtract $-\log(3)u$
$(5)$: exponentiate and negate
Therefore, applying Lambert W to $(5)$ yields
$$
\begin{align}
x
&=u-24\\
&=-\frac{\operatorname{W}\left(-3^{-24}\log(3)\right)}{\log(3)}-24\tag6
\end{align}
$$
There are two real branches of Lambert W for negative arguments. This leads to two real solutions:
$$
x=-23.9999999999964592938385140774\tag7
$$
and
$$
x=3\tag8
$$
A: Rearranging the equation gives the more reasonable-looking equation $$x + 24 = 3^x .$$
One can now show using straightforward facts from calculus (the strict convexity of exponentials that these are at most two solutions. Then, by inspection: (1) $x = 3$ is a solution, and (2) there is a second solution very close to $x = -24$, as rearranging gives $x + 24 = 3^x$ and $3^{-24} \approx 0 = (-24) + 24$.
One can write these solutions (as well as solutions of similar equations for which there isn't a nice [say, integral] solution like $x = 3$ in our example) "explicitly" using the so-called Lambert W function, but it's not always immediately apparent when a solution written this way has a nicer expression.
A: \begin{gather*}
(x+24)^{\frac{1}{x}} \ =\ 3\ \ \ \ \ \ \ \ \ \ \ \frac{1}{x} ln(x+24)\ =\ ln3\\
\\
ln(x\ +\ 24)\ \ \ =\ \ \ xln3\ \ \ =\ \ \ ln3^{x}\\
\\
e^{ln(x+24)} \ =\ e^{ln3^{x}} \ \ \ \ \ \ \ \ \ \ \ \ \ x+24\ =\ 3^{x}
\end{gather*}
\begin{gather*}
Now\ Graph\ f(x)\ =\ x\ +\ 24\ and\ g(x)\ =\ 3^{x} \ on\ the\ same\ graph\\
and\ the\ x\ coordinate\ of\ the\ point\ where\ the\ graphs\ intersect\ is\ \\
the\ solution:\ x\ =\ 3 \ and\ x \sim -24
\end{gather*}

A: It is perhaps instructive to see how one solution of $x = 3$ arises from the Lambert W function solution as given by @robjohn.
As was shown, the solution to the equation is 
$$x = -24 - \frac{\text{W}_\nu (-3^{-24} \ln 3)}{\ln 3}. \tag1$$
Here $\nu$ denotes the two real branches for the Lambert W function. When $\nu = 0$ we have the principal branch while when $\nu = -1$ we have the second real branch.
A know simplification rule for the Lambert W function is
$$-\ln t = \begin{cases}
\text{W}_0 \left (-\dfrac{\ln t}{t} \right ), \quad 0 < t \leqslant e,\\[2ex]
\text{W}_{-1} \left (-\dfrac{\ln t}{t} \right ), \quad t \geqslant e.
\end{cases}$$
Now consider the term containing the Lambert W function in (1). We can write this as
\begin{align*}
\text{W}_\nu (-3^{-24} \ln 3) & = \text{W}_\nu \left (-\frac{\ln 3}{3^{24}} \right )\\
&= \text{W}_\nu \left (-\frac{27 \cdot \ln 3}{27 \cdot 3^{24}} \right )\\
&= \text{W}_\nu \left (-\frac{\ln (3^{27})}{3^3 \cdot 3^{24}} \right )\\
&= \text{W}_\nu \left (-\frac{\ln (3^{27})}{3^{27}} \right ).
\end{align*}
Now as $t = 3^{27} > e$, from the simplification rule given above we see the secondary real branch ($\nu = -1$) for the expression containing the Lambert W function simplifies. Here
$$\text{W}_{-1} \left (-\frac{\ln (3^{27})}{3^{27}} \right ) = - \ln (3^{27}).$$
So the two solutions to the equation are
$$ x = \begin{cases}
-24 - \dfrac{\text{W}_0 (-3^{-24} \ln 3)}{\ln 3}, \quad \text{and}\\[2ex]
3.
\end{cases}$$
