Solving $5\times9^x-8\times15^x+3\times25^x>0$

Question

Question: Solve the inequality:

$5\times9^x-8\times15^x+3\times25^x>0$

So far I have managed to factorise the inequality:

$(5\times3^x-3\times5^x)(3^x-5^x)>0$

From here on, I am stuck.

I would very much appreciate it if you could answer this exam question and lay the working out as if it were an exam. I have an answer from the solutions book so can tell you if your answer is correct using that. Many thanks.

Answer 1

You may solve your inequality by transforming it into a sign problem for a quadratic function dividing by $15^x= 3^x\cdot 5^x$:

$$5\times9^x-8\times15^x+3\times25^x>0 \Leftrightarrow 5\left(\frac 35\right)^x - 8 + 3\left(\frac 53\right)^x>0$$

Setting $t = \left(\frac 35\right)^x$ you only need to solve

$$5t+\frac 3t - 8 >0 \text{ for } t>0$$

or equivalently $$5t^2 - 8t + 3 = (5t-3)(t-1)>0$$

So, you get either

$$1<t= \left(\frac 35\right)^x \Leftrightarrow \boxed{x<0}$$

or

$$\frac 35 > t =\left(\frac 35\right)^x \Leftrightarrow \boxed{x>1}$$

Answer 2

The first factor is positive when $$5 \cdot 3^x - 3 \cdot 5^x > 0$$ divide by $15$ to get $$3^{x-1} - 5^{x-1} > 0$$ $$3^{x-1} > 5^{x-1}$$ $$\frac{3^{x-1}}{5^{x-1}} >1$$ $$\left( \frac{3}{5}\right)^{x-1} >\left( \frac{3}{5}\right)^{0}$$ $$x-1<0$$ $$x<1$$

In a similar way you can show that the second factor is positive when $x<0$.

Hence the solution is $$x<0 \qquad \mbox{ or } \qquad x>1$$

Answer 3

When both factors are positive, the intersection of $x < 0$ and $x < 1$ is $x < 0$. But when both factors of $(5\times3^x-3\times5^x)(3^x-5^x)$ are negative, their product is still positive, so we must consider this case as well.

$5\times3^x-3\times5^x < 0$ gives $\left( \frac{3}{5}\right)^{x-1} < \left( \frac{3}{5}\right)^{0}$ like before. However, since $3/5 < 1$, the $y$-values are getting smaller as the $x$-values increase. Drawing a sketch of the exponential function and the curve, the inequality sign flips to give $x - 1 > 0, x > 1$.

Similarly, $3^x - 5^x < 0$ leads to $3^x < 5^x$ or $\left( \frac{3}{5}\right)^x < \left( \frac{3}{5}\right)^{0}$, and the inequality sign flips again, thus $x > 0$. Since both factors need to be negative for the product to be positive, we need to take the intersection of $x > 1, x > 0$ which is $x > 1$.

Altogether, we have $x < 0$ and $x > 1$.