An algebraic solution for $\log_3(x+1)+\log_2(x)=5$ The logarithmic equation 
$$\log_3(x+1)+\log_2(x)=5$$
has an obvious solution, namely $x=8$. However, I can't seen to find an algebraic demonstration/deduction of this fact. This has been an "unsolvable" problem for me since my early days in elementary/middle school. Any solution not relying on inspection would be appreciated.
EDIT: Driven by @TobyMak's comments: The main issue here is that this problem was supposed to be solved by a middle school student. Using analysis and knowing beforehand that $x=8$ solves the equation does the job. I would like to know if there are finite algebraic steps which lead to the result.  
 A: $$\log_3(x+1)+\log_2 x=5$$
$$\log_3(x+1)=5-\log_2 x$$
$$x+1=3^{5-\log_2 x}$$
For integer solutions for x we must have:
$$5-\log_2 x>0$$
⇒ $$\log_2 x<5$$
Therefore we must check numbers 4, 3,2, 1 which gives:
$\log_2 x= 1, 2, 3, 4$
⇒$x=2, 4, 8, 16$
These solution must also satisfy the initial equation; corresponding values are:
$\log_3 (x+1)=5-\log_2x=5-1=4,5-2= 3,5-3= 2,5-4= 1$
⇒ $x+1= 3^4=81, 3^3=27, 3^2=9, 3^1=3$
⇒$x= 80, 26, 8, 2$
The only common solution is 8. 
A: Let $f(x)=\log_3(x+1)+\log_2x.$
Thus, since $f$ increases, our equation has one root maximum.
$8$ is a root, which says that it's an unique root and we are done.
A: I tried to think of how a middle schooler might solve this:
Let $x=2^a$.  Then we have
$$\log_3 (2^a+1) + a = 5.$$ So that
$$2^a+1 = 3^{5-a}.$$
Multiply by $3^a$ to get
$$6^a+3^a = 3^5 = 243.$$
If the student knows that $6^3 = 216,$  he knows that $a$ is pretty close to $3$, which does in fact work, giving $x=8.$
A: I don't know if this is what you want, but I hope it helps.
$\log_2(x)=a$ means that $2^a=x$
$\log_3(x+1)=b$ means that $3^b=x+1$
To get $a+b=5$ where $3^b$ is after $2^a$ by $+1$ only (somehow consecutive numbers), then it is most likely that $a$ and $b$ are integers. (This step isn't reasonable actually, it is just "looks like so")
Now if $a$ and $b$ are integers, the ofcourse $x$ is also an integer, so now you'll be searching for $2$ consecutive numbers where the first is a power of $2$ and the second is a power of $3$, and there is the condition that $a+b=5$ so if you start trying natural numbers, the first 2 numbers that satisfy the above conditions are $8=2^3$ and $9=3^2$, and $2+3=5$, so your only answer is $x=8$.
