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

Question

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.

Answer 1

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

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

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

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.

Answer 2

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 $x=8$.

Answer 3

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.

Answer 4

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 $x=8.$