I am struggling with a logarithmic equation where I am required to find integer solutions for a such that x is also an integer and greater than 4. The equation is given as:

$a = \log_2\left(\frac{9 \cdot 2^x - 112}{2^x - 13}\right) - 1$

To clarify, I am looking for the values of x that make a an integer where x > 4.

I understand that for a to be an integer, the argument of the log base 2 (i.e., $\frac{9 \cdot 2^x - 112}{2^x - 13}$) must be a power of 2. The constants in the numerator and denominator, along with the subtraction of 1, make it non-trivial to find such x.

Could anyone provide insight or a method to determine the possible integer values of x that satisfy the given conditions? Any assistance or suggestions on how to approach this would be greatly appreciated.

Thank you for your time and help!

  • 2
    $\begingroup$ $\frac{9\cdot 2^x - 112}{2^x-13} = 9+\frac{5}{2^x-13}$. This is never an integer for $x>4$, let alone a power of 2. $\endgroup$
    – D S
    Feb 25 at 10:06

2 Answers 2


For $a$ to be an integer $f(x)=\log_2\left(\frac{9\cdot2^x-112}{2^x-13}\right)$ must be an integer. You can graph this and look for integer solutions:

$\log_2\left(\frac{9\cdot2^x-112}{2^x-13}\right)$ has a vertical asymptote at $9\cdot2^{x}-112$ or $x=\frac{\log_2(112/9)}{\log_2(2)}\approx 3.64$.

The limit of $\log_2\left(\frac{9\cdot2^x-112}{2^x-13}\right)$ as $x\to +\infty$ is $\frac{\log_2(9)}{\log_2(2)}\approx 3.17$.

The limit of $\log_2\left(\frac{9\cdot2^x-112}{2^x-13}\right)$ as $x\to -\infty$ is $\frac{\log_2(112/13)}{\log_2(2)}\approx 3.107$.

Furthermore, by taking the derivative, one can show the function is strictly decreasing everywhere it is continuous.

Evaluating at some integer points:

$f(4)=3.415037$ is not an integer. As $f(x)$ is stricly decreasing to $3.17$, there will be no integer solutions for $x\geq4$.

$f(2)=3.078002$ and as $\lim_{x\to -\infty}f(x)=3.107$, there are no integer solutions for $x\leq 2$.

The last point to check is $x=3$, which evaluates to $f(3)=3$, the only solution. There are no solutions for $x>4$.

  • $\begingroup$ I'm not sure how to solve this algebraically... $\endgroup$ Feb 25 at 10:16

We have equivalently $2^{a+1}=\dfrac{9 \cdot 2^x - 112}{2^x - 13}$ and making the division,$$2^{a+1}=9+\dfrac{5}{2^x-13}$$ For $x\gt4$ one has $2^x-13\ge19$ so there are no integer solutions.

(We can note that $x=3$ would be a solution but we have the restriction $x\gt4$).


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