Question

$\left(\frac{3^{x-2}}{3^x-2^x}\right)>1+\left(\frac{2}{3}\right)^x$

Answer 1

$\left(0, \log_{2/3}\left(\frac{2\sqrt{2}}{3}\right)\right)$

Answer 2

The given inequality is $\left(\frac{3^{x-2}}{3^x-2^x}\right)>1+\left(\frac{2}{3}\right)^x$.

First, simplify the left side: $\frac{3^{x-2}}{3^x-2^x} = \frac{3^x \cdot 3^{-2}}{3^x(1 - \frac{2^x}{3^x})} = \frac{3^{-2}}{1 - (\frac{2}{3})^x} = \frac{1/9}{1 - (\frac{2}{3})^x}$.

Let $y = (\frac{2}{3})^x$. Since $2/3 > 0$, $y > 0$ for all real $x$. The inequality becomes: $\frac{1/9}{1-y} > 1+y$.

The denominator $1-y$ cannot be zero, so $y \neq 1$. This means $(\frac{2}{3})^x \neq 1$, which implies $x \neq 0$.

Rearrange the inequality: $\frac{1/9}{1-y} - (1+y) > 0$ $\frac{1/9 - (1+y)(1-y)}{1-y} > 0$ $\frac{1/9 - (1-y^2)}{1-y} > 0$ $\frac{1/9 - 1 + y^2}{1-y} > 0$ $\frac{y^2 - 8/9}{1-y} > 0$ $\frac{(y - \sqrt{8}/3)(y + \sqrt{8}/3)}{1-y} > 0$ $\frac{(y - 2\sqrt{2}/3)(y + 2\sqrt{2}/3)}{1-y} > 0$.

Since $y = (2/3)^x > 0$, the term $(y + 2\sqrt{2}/3)$ is always positive. We can divide the inequality by this term without changing the sign: $\frac{y - 2\sqrt{2}/3}{1-y} > 0$.

This inequality holds if and only if the numerator and denominator have the same sign.

Case 1: $y - 2\sqrt{2}/3 > 0$ and $1-y > 0$. $y > 2\sqrt{2}/3$ and $y < 1$. So, $2\sqrt{2}/3 < y < 1$.

Case 2: $y - 2\sqrt{2}/3 < 0$ and $1-y < 0$. $y < 2\sqrt{2}/3$ and $y > 1$. Since $2\sqrt{2}/3 = \sqrt{8/9} < \sqrt{1} = 1$, the condition $y > 1$ and $y < 2\sqrt{2}/3$ has no solution.

So the only solution for $y$ is $2\sqrt{2}/3 < y < 1$.

Substitute back $y = (\frac{2}{3})^x$: $2\sqrt{2}/3 < \left(\frac{2}{3}\right)^x < 1$.

This inequality can be split into two parts:

1. $\left(\frac{2}{3}\right)^x < 1$ Since $1 = (\frac{2}{3})^0$ and the base $2/3$ is between 0 and 1, the function $(\frac{2}{3})^x$ is decreasing. $(\frac{2}{3})^x < (\frac{2}{3})^0 \implies x > 0$.

2. $2\sqrt{2}/3 < \left(\frac{2}{3}\right)^x$ Let $x_0 = \log_{2/3}(2\sqrt{2}/3)$. This is the value of $x$ for which $(\frac{2}{3})^x = 2\sqrt{2}/3$. So the inequality is $(\frac{2}{3})^{x_0} < \left(\frac{2}{3}\right)^x$. Since the base $2/3$ is between 0 and 1, the inequality sign flips when comparing the exponents: $x_0 > x$.

Combining the two conditions $x > 0$ and $x < x_0$, we get $0 < x < x_0$.

The value of $x_0$ is $\log_{2/3}(2\sqrt{2}/3)$. We can write $2\sqrt{2}/3 = 2^{3/2} \cdot 3^{-1}$. $x_0 = \log_{2/3}(2^{3/2} \cdot 3^{-1}) = \frac{\ln(2^{3/2} \cdot 3^{-1})}{\ln(2/3)} = \frac{\frac{3}{2}\ln 2 - \ln 3}{\ln 2 - \ln 3}$.

The solution set for $x$ is $(0, \log_{2/3}(2\sqrt{2}/3))$.

The final answer is $\left(0, \log_{2/3}\left(\frac{2\sqrt{2}}{3}\right)\right)$.

Explanation of the solution:

1. Simplify the given inequality by expressing terms with a common base $(2/3)^x$.
2. Introduce a substitution $y = (2/3)^x$ to transform the inequality into a rational inequality in terms of $y$.
3. Solve the rational inequality for $y$, considering the constraint $y > 0$ (since $y = (2/3)^x$).
4. Convert the solution for $y$ back into terms of $x$ using $y = (2/3)^x$.
5. Solve the resulting exponential inequalities for $x$, taking into account that the base $2/3$ is less than 1, which reverses the inequality direction when comparing exponents.
6. Combine the solutions from the exponential inequalities to find the final range for $x$. The logarithm base $2/3$ is used to express the boundary value.

Answer: The solution is the interval $\left(0, \log_{2/3}\left(\frac{2\sqrt{2}}{3}\right)\right)$. This corresponds to the range $0 < x < \log_{2/3}\left(\frac{2\sqrt{2}}{3}\right)$.