How to find x if $\log_2(x)\cdot\log_2(x+2)=4$?

Question

I get this question for a high school book.(I can't remember that book.)

I think question actually should be $\log_2(x)+\log_2(x+2)=4?$ which high school student can solve. But I want to know that "Is it possible to solve that wrong printing question by using higher knowledge?"

Answer 1

Let $x=2^k$

$$\log_2(x)\cdot\log_2(x+2)=4\\\iff \log_2(2^k)\cdot\log_2(2^k+2)=4 \\\iff \log_2(2^k+2)=\frac4k \\\iff 2^k+2=16^\frac1k$$

from here we can easily show by IVT that exactly one solution exists and evaluate it numerically, indeed

• $f(k)=2^k+2$ is strictly increasing and $f(1)=4$ and $f(4)=18$
• $g(k)=16^\frac1k$ is strictly decreasing and $g(1)=16$ and $g(4)=2$

thus an unique solution exists for $k\in(1,4)$.

Answer 2

This was for the supposed question. Please ignore.

Note that $\log_2 16=4$ The equation can be therefore written as $\log_2 x + \log_2 (x+2)= \log_2 16 $

So we have $\log_2 (x^2+2x)= \log_2 16\\x^2+2x=16$

Using the quadratic formula the positive solution is $\sqrt{17} -1$