Review question

If we arrange $2^x, 2x$ and $x^2$ by size, what orders are possible? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5156

Solution

Which of the following have no real solutions?

(i) $2x<2^x<x^2\quad$ (ii) $x^2<2x<2^x\quad$ (iii) $2^x<x^2<2x\quad$
(iv) $x^2<2^x<2x\quad$ (v) $2^x<2x<x^2\quad$ (vi) $2x<x^2<2^x$

A (i) and (iii),$\quad$ B (i) and (iv),$\quad$ C (ii) and (iv),$\quad$ D (ii) and (v),$\quad$ E (iii) and (v)

It helps here to think of the graphs $y = 2x$, $y = 2^x$ and $y = x^2$.

The curves $y = 2x$ and $y = x^2$ meet at the origin.

When $x = 1$, $2^x = 2x$.

When $x = 2$, $2^x = x^2=2x$.

When $x = 4$, $2^x = x^2$.

From these facts, we can sketch the three curves and note what order they must be in between their intersections.

The red curve is $y = 2x$, the blue curve is $y=x^2$ and the green curve is $y=2^x$.

We can see that the possible orders are
red < green < blue, red < blue < green, blue < red < green, blue < green < red.

So the green curve cannot be the bottom one, and that rules out (iii) and (v). The answer is E.