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.
Best Answer
$$\log_3(x+1)+\log_2 x=5$$
$$\log_3(x+1)=5-\log_2 x$$
$$x+1=3^{5-\log_2 x}$$
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.