Solving logarithmic equations including x Let $$\log_3(x-2) = 6 - x$$
It's obvious drawing the graphs of the two functions that the only solution is $x=5$. But this is not really a proof, rather than observation.
How do you prove it algebraically? 
 A: We can solve this equation for $x$ in closed form as follows.
Write $\log_3 (x-2)=6-x$ as $\frac{\log (x-2)}{\log 3}=6-x$.  Then we have
$$x-2=e^{\log 3[4-(x-2)]}=3^4e^{-(\log 3)\,(x-2)}\tag 1$$
Thus, multiplying $(1)$ by $\log 3$ and rearranging terms gives
$$(\log 3)(x-2)e^{(\log 3)(x-2)}=3^4\,\log 3$$
Recalling that Lambert's W function is defined as $z=W(z)e^{W(z)}$ gives
$$\begin{align}
x&=2+\frac{W(3^4\,\log 3)}{\log 3}\\\\
&=2+3 \\\\
&=5
\end{align}$$
as was to be shown!

NOTE:
To show that $W(3^4\log 3)=3\log 3$ we use the interesting property of the W function that when $W(z)=x\log x$, $z=x^{x+1}\log x$.  So, here we take $x=3$ and note that $W(3^4\log 3)=3\log 3$ ... as expected!
A: Finding the solution $x=5$ is just a matter of luck, try and error, approximation methods or just observation. 
I guess that you are looking for a way to show that there are no more solutions. Just nothe that the function
$$f(x)=\log_3(x-2)$$
is strictly increasing and the function
$$g(x)=6-x$$ is strictly decreasing. Therefore
$$f(x)<1<g(x)$$
for $2<x<5$ and
$$g(x)<1<f(x)$$
for $x>5$.
A: Since one has $x+\log_3(x-2)=6$, let $f(x)=x+\log_3(x-2)$. Note that $x\gt 2$.
Now, 
$$f'(x)=1+\frac{1}{(x-2)\ln 3}\gt 0.$$
Hence, since $y=f(x)$ is increasing for $x\gt 2$, we know that $f(x)=6$ has at most one real solution (you know that it is $x=5$).
A: Unfortunately there isn't a straightforward way to show this algebraically.
We know by inspection that $x=5$ is a solution since $$\log_3(5-2)=\log_3(3) = 1 = 6-5.$$
With a bit of calculus, you can demonstrate that these would not intersect again. This uses what is called the Rolle's Theorem.
Let $f(x) = \log_3(x-2) - 6 + x$. If $y$ satifies your equation, then $f(y)=0$.
We know that $f(5)=0$ for example.
Rolle's Theorem says that if there is another solution $y$, then there is a point $\xi$ in the interval $[5,y]$ (we are assuming $y>5$ here, but the same argument works for $y<5$) for which $f'(\xi)=0$.
However $f'(x) = \frac{1}{(x-2)\ln(3)} + 1 = \frac{1+(x-2)\ln(3)}{(x-2)\ln(3)}$ which means $1+(\xi-2)\ln(3) =0$ or $\xi =2-\ln(3)^{-1}$. This is outside the domain of $\log_3(x-2)$, which is a contradiction.
Thus the only solution is $x=5$.
A: You don't: other than in some special cases equations involving logarithms, exponentials, or trigonometric functions (to name a few) don't have algebraic solutions. You can prove it analytically by showing that the LHS is strictly increasing with a zero at $x = 3$, while the RHS is strictly decreasing and positive at $x = 3$.
