\(\sqrt{81^{1\:-\:2x}} \geq 27^{2\:-\:x}\)

\(81^{\frac{1\:-\:2x}{2}} \geq 3^{3(2\:-\:x)}\)

\(3^{\cancelto{2}{4}\cdot\frac{1\:-\:2x}{\cancel{2}}} \geq 3^{3(2\:-\:x)}\)

\(3^{2(1\:-\:2x)} \geq 3^{3(2\:-\:x)}\)

\(2(1\:-\:2x) \geq 3(2\:-\:x)\)

\(2 \:-\: 4x \geq 6 \:-\: 3x\)

\(2\:-\:6 \geq -3x + 4x\)

\(-4 \geq x\)

\(x \leq -4\)

Jadi solusi pertidaksamaannya adalah \(x \leq -4\)

\(\left(\dfrac{1}{2}\right)^{x^2 \:-\:3x \:-\: 1} < \left(\dfrac{1}{2}\right)^{3(x^2 \:-\: 12 x + 5)}\)

Karena bilangan pokok (basis)  \(0 < \dfrac{1}{2} < 1\) maka tanda pertidaksamaan dibalik

\(x^2 \:-\: 3x \:-\: 1 > 3(x^2 \:-\: 12 x + 5)\)

\(x^2 \:-\: 3x \:-\: 1 > 3x^2 \:-\: 36x + 15\)

\(0 > 3x^2\:-\:x^2 – 36x + 3x + 15 + 1\)

\(0 > 2x^2 \:-\: 33x + 16\:\:\:\:\:\color{cyan}\text{faktorkan}\)

\(0 > (2x\:-\:1)(x \:-\: 16)\)

\(\text{Pembuat nol:}\)

\(2x \:-\: 1 = 0 \rightarrow x = \dfrac{1}{2}\)

\(x \:-\: 16 = 0 \rightarrow x = 16\)

\(5^{2x} \:-\: 30\cdot 5^x + 125 > 0\)

\(\text{Misal: } p = 5^x\)

\(p^2\:-\: 30p + 125 > 0\:\:\:\:\:\color{cyan}\text{faktorkan}\)

\((p \:-\:5)(p \:-\: 25) > 0\)

\(\text{Pembuat nol:}\)

\(p \:-\:5 = 0 \rightarrow p = 5\)

\(p \:-\:25 = 0 \rightarrow p = 25\)

\(2^{x + 2} \:-\:\dfrac{40}{2^x} \:-\: 27 < 0\)

\(2^2\cdot 2^{x}\:-\: \dfrac{40}{2^x} \:-\: 27 < 0\)

\(\text{Misal: } p = 2^x\)

\(4p \:-\: \dfrac{40}{p} \:-\:27 < 0\:\:\:\:\:\color{cyan}\text{samakan penyebut}\)

\(\dfrac{4p^2 \:-\:40 \:-\: 27p}{p} < 0\)

\(\dfrac{4p^2 \:-\:27p \:-\: 40}{p} < 0\:\:\:\:\:\color{cyan}\text{faktorkan}\)

\(\dfrac{(4p + 5)(p \:-\: 8)}{p} < 0\)

\(\text{Pembuat nol:}\)

\(4p + 5 = 0 \rightarrow p = -\dfrac{5}{4}\)

\(p\:-\: 8 = 0 \rightarrow p = 8\)

\(p \neq 0\)

Ubah pecahan desimal menjadi bentuk pecahan biasa yang sederhana

\(0,16 = \dfrac{16}{100} = \dfrac{4}{25} = \left(\dfrac {2}{5}\right)^2\)

\(0,0256 = \dfrac{256}{10000} = \dfrac{16}{625} = \left(\dfrac{2}{5}\right)^4\)

\((0,16)^{x^2 \:-\: 3x +1} \leq (0,0256)^{x^2\:-\: 1,5}\)

\(\left(\dfrac {2}{5}\right)^{2(x^2 \:-\: 3x +1)} \leq \left(\dfrac{2}{5}\right)^{4(x^2 \:-\: 1,5)}\)

Karena bilangan pokok (basis)  \(0 < \dfrac{2}{5} < 1\) maka tanda pertidaksamaan dibalik

\(2(x^2 \:-\:3x +1) \geq 4(x^2 \:-\: 1,5)\)

\(2x^2 \:-\: 6x + 2 \geq 4x^2 \:-\: 6\)

\(0 \geq 4x^2 \:-\: 2x^2 + 6x \:-\: 6 \:-\: 2\)

\(0 \geq 2x^2 + 6x \:-\:8\:\:\:\:\:\color{cyan}\text{bagi kedua ruas dengan 2}\)

\(0 \geq x^2 + 3x \:-\: 4\:\:\:\:\:\color{cyan}\text{faktorkan}\)

\(0 \geq (x \:-\: 1)(x + 4)\)

\(\text{Pembuat nol: }\)

\(x \:-\:1 = 0 \rightarrow x = 1\)

\(x + 4 = 0 \rightarrow x = -4\)