Question

(1/3) ^ x > (1/3) ^ - 2​❗ТЕРМІНОВО ❗

Answer 1

$\displaystyle\bigg(\frac{1}{3} \bigg)^\bigg x > \bigg(\frac{1}{3} \bigg)^\bigg {-2}$

Т.к. 1/3<1, то знак меняется, следовательно

$x < -2\\\\x\in (-\infty;-2)$

Answer 2

To solve the inequality $(1/3)^x > (1/3)^{-2}$, we can start by simplifying both sides:

$(1/3)^x > (1/3)^{-2}$

Since $(1/3)^{-2} = 3^2 = 9$, the inequality becomes:

$(1/3)^x > 9$

Next, we can express $9$ as a power of $\frac{1}{3}$ since $\frac{1}{3}^2 = \frac{1}{9}$:

$(1/3)^x > (1/3)^{-2}$

$(1/3)^x > (1/3)^{2}$

$(1/3)^x > 1/9$

Now, we have:

$(1/3)^x > 1/9$

To solve this inequality, we can take the logarithm base $\frac{1}{3}$ of both sides:

$x < \log_{1/3}(1/9)$

$x < \log_{1/3}(1/3^2)$

$x < \log_{1/3}(1/3^{2})$

$x < \log_{1/3}(1/3^{2})$

$x < \log_{1/3}(1/9)$

$x < 2$

So, the solution to the inequality is $x < 2$.