Logarithmic inequation $\frac{3^{x+1}}{2^{x+3}}+2\leq \frac{2^{x-1}}{3^{x-1}}$

Question

Can somebody help me with this problem?

Solve inequation

$$\frac{3^{x+1}}{2^{x+3}}+2\leq \frac{2^{x-1}}{3^{x-1}}$$

I am trying to solve it, but I am stuck here...

$$\dfrac{3^{2x}-4^{2x}+\frac{16}3\times 6^x}{\frac83\times6^x}\leq 0$$

Answer 1

Multiplying through by $\frac{3^{x-1}}{2^{x-1}}$ gives: $$\frac{3^{2x}}{2^{2x+2}} + \frac{2\cdot 3^{x-1}}{2^{x-1}} \leq 1$$ or equivalently, $$\frac{1}{4}\left(\frac{3}{2}\right)^{2x} + \frac{4}{3}\left(\frac{3}{2}\right)^x-1 \leq 0.$$ Let $y=\left(\frac{3}{2}\right)^x$. We want to solve $\frac{1}{4}y^2+\frac{4}{3}y-1 \leq 0$. Using the quadratic formula or factoring, we find the roots of this equation are $y=-6$ and $y=\frac{2}{3}$, so $\frac{1}{4}y^2+\frac{4}{3}y-1 \leq 0$ whenever $y \in [-6,\frac{2}{3}]$.

So finally, we need to solve the inequalities $-6 \leq y \leq \frac{2}{3}$, i.e., $-6 \leq \left(\frac{3}{2}\right)^x \leq \frac{2}{3}$. Since $\left(\frac{3}{2}\right)^x >0$ for all $x$, the lower bound always holds. Thus the inequality is reduced to solving $\left(\frac{3}{2}\right)^x \leq \frac{2}{3}$. Can you continue from here?

Answer 2

Since everything in sight is continuous, we should first solve the corresponding equality. Clear the fractions and put everything on one side to get

$$2^{2x+2} -2\cdot3^{x-1}2^{x+3} - 3^{2x} = 0.$$

Pull out the odd powers:

$$4\cdot 2^{2x} - 16\cdot 2^x \frac{1}{3} 3^x -3^{2x}=0.$$

Multiply by $3$ and let $a=2^x$ and $b=3^x$:

$$12a^2 - 16ab - 3b^2 = 0.$$

This factors

$$(6a-b)(2a-3b) = 0.$$

This gives two easy equations in $x$ leading to two solutions. From there the inequality is easy.

Answer 3

First, your inequality, $$ \frac{3^{2x}-4^{2x}+\frac{16}{3}\times 6^x}{\frac{8}{3}\times 6^x} \leq 0 \text{,} $$ has an always positive denominator on the left-hand side, so the denominator cannot make the left-hand side zero or change its sign. Hence, you can reduce to $$ 3^{2x}-4^{2x}+\frac{16}{3}\times 6^x \leq 0 \text{.} $$

However, much better is to note that in your original inequality, you very nearly have the same expressions involving exponents. Let's see if we can make this matching exact. $$ \frac{9}{16} \left( \frac{2^{x-1}}{3^{x-1}} \right)^{-1} + 2 \leq \frac{2^{x-1}}{3^{x-1}} \text{.} $$ Let $u = \frac{2^{x-1}}{3^{x-1}}$ so that the inequality is $$ \frac{9}{16u} + 2 \leq u \text{,} $$ which is much easier to solve. You will get two intervals for $u$. But $\frac{2^{x-1}}{3^{x-1}}$ is always positive, so we can restrict the intervals by intersecting with $(0,\infty)$ to get solutions for $u$ that correspond to valid values of $x$.