Review question

# When does this inequality have a finite number of integer solutions? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6414

## Solution

Let $a$, $b$, $c$ be positive numbers. There are finitely many positive whole numbers $x$, $y$ which satisfy the inequality $a^x > cb^y$ if

1. $a>1$ or $b<1$.

2. $a<1$ or $b<1$.

3. $a<1$ and $b<1$.

4. $a<1$ and $b>1$.

Taking logarithms of both sides we have $a^x > cb^y \quad\iff\quad \log(a^x)> \log(cb^y)$ (so long as the base of the logarithm is greater than one, as $\log x$ is then an increasing function).

This can be rewritten as \begin{align*} x\log a &> \log c+y\log b\\ \iff\quad y\log b &< x\log a - \log c \end{align*}

The corresponding equation is of a straight line in the $x-y$ plane. To find the gradient of this line we want to divide through by $\log b$ but we must be careful as this could be negative.

We have either $\text{Case 1: }\quad y < x\frac{\log a}{\log b} - \frac{\log c}{\log b},\quad\text{if } \log b > 0$ or $\text{Case 2: }\quad y > x\frac{\log a}{\log b} - \frac{\log c}{\log b},\quad\text{if } \log b < 0.$

To have a finite number of positive solutions the inequality must correspond to the region below a straight line with negative gradient.

Case 2 is a region above a line, so we need only consider case 1, $\log b > 0$. We also require the gradient of the line, $\dfrac{\log a}{\log b}$, to be negative.

Hence, $\log b > 0$ and $\log a < 0$. In other words, $b > 1$ and $a < 1$.

So the answer is (d).

Why does it not matter whether the $y$-intercept of the straight line is positive or negative?