# Finding integer solutions for a logarithmic equation with constraints

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!

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

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 $$x>4$$.

• I'm not sure how to solve this algebraically... 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 $$x\gt4$$).