An algebraic solution for $\log_3(x+1)+\log_2(x)=5$

$$\log_3(x+1)+\log_2 x=5$$

$$\log_3(x+1)=5-\log_2 x$$

$$x+1=3^{5-\log_2 x}$$

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.

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.

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 $x=8.$