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

\(a>1\) or \(b<1\).

\(a<1\) or \(b<1\).

\(a<1\) and \(b<1\).

\(a<1\) 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?