Inequaciones exponenciales - problemas y soluciones

Solución:
$5^{x^2+3} \le 5^{4x}$

$x^2+3 \le 4x$
$x^2-4x+3 \le 0$
$x^2-4x+3 = 0$

$D=4$
$X_{1}=1$
$X_{2}=3$

$x^2-4x+3 \le 0 \Leftrightarrow x \in [1,3]$

Respuesta: $x \in [1,3]$

Solución:
$3^{\frac{x-3}{3x-2}} \le 3^{1}$

$\frac{x-3}{3x-2} \le 1$

$\frac{x-3}{3x-2} -1 \le 0$

$\frac{x-3-1(3x-2)}{3x-2} \le 0$

$\frac{x-3-3x+2}{3x-2} \le 0$

$\frac{-2x-1}{3x-2} \le 0 \qquad /\cdot -1$

$\frac{2x+1}{3x-2} \ge 0$

$2x+1 = 0 \Leftrightarrow 2x = -1 \Leftrightarrow x = -\frac {1}{2}$
$3x-2=0 \Leftrightarrow 3x = 2 \Leftrightarrow x = \frac {2}{3}$

$\frac{2x+1}{3x-2} \ge 0 \Leftrightarrow x \in (-\infty,-\frac{1}{2}] \cup (\frac{2}{3},+\infty) $

Respuesta: $x \in (-\infty,-\frac{1}{2}] \cup (\frac{2}{3},+\infty)$

Solución:
$0 < \frac{3}{4} < 1 \Rightarrow$

$\left(\frac{3}{4} \right)^{2-x} > \left(\frac{3}{4} \right)^{-\frac{1}{2}} \Leftrightarrow$

$2-x < -\frac{1}{2} \qquad /.2$

$4-2x < -1$
$5<2x$

$x>\frac{5}{2}$

Respuesta: $x \in (\frac{5}{2},+\infty)$

Solución:
$\left(\frac{1}{3} \right)^{2x^2-5} < \left(\frac{1}{9} \right)^{\frac{3}{2}}$

$\left(\frac{1}{3} \right)^{2x^2-5} < \left(\left(\frac {1}{3}\right)^2 \right)^{\frac{3}{2}}$

$\left(\frac{1}{3} \right)^{2x^2-5} < \left(\frac{1}{3} \right)^{2 \times \frac{3}{2}}$

$\left(\frac{1}{3} \right)^{2x^2-5} < \left(\frac{1}{3} \right)^{3}$

$0 < \frac{1}{3} < 1$

$\left(\frac{1}{3} \right)^{2x^2-5} < \left(\frac{1}{3} \right)^{3} \Leftrightarrow$

$2x^2-5 > 3$
$2x^2-8 > 0 \qquad / :2$
$x^2-4 > 0$
$(x-2)(x+2) > 0 \Leftrightarrow x \in (-\infty,-2) \cup (2,+\infty)$

Respuesta: $x \in (-\infty,-2) \cup (2,+\infty)$

Solución:
$0 < 0{,}4 < 1$
$\Rightarrow$

${0{,}4}^{3x-1} \ge {0{,}4}^{x+1} \Leftrightarrow 3x-1 \le x+1$
$3x-1 \le x+1$
$2x \le 2$
$x \le 1$

Respuesta: $x \in (-\infty,1]$

Solución:
${5}^{x^2-6x+8} < 1$

${5}^{x^2-6x+8} < {5}^{0}$

$x^2-6x+8 < 0$
$x^2-6x+8 = 0$

$D=4$
$x_{1}=4$ ; $x_{2}= 2$
$\Rightarrow$
$x^2-6x+8 < 0 \Leftrightarrow x\in (2,4)$

Respuesta: $x\in (2,4)$

Solución:
$\left( \frac {3}{7} \right)^{\frac {x^2-2x}{x^2}} \ge 1$

Dominio: $x^2 \ne 0 \Leftrightarrow x \Leftrightarrow 0$

$\left( \frac {3}{7} \right)^{\frac {x^2-2x}{x^2}} \ge \left( \frac {3}{7} \right)^{0}$

$0 < \frac {3}{7} < 1 \Rightarrow$

$\left( \frac {3}{7} \right)^{\frac {x^2-2x}{x^2}} \ge \left( \frac {3}{7} \right)^{0} \Leftrightarrow \frac {x^2-2x}{x^2} \le 0$

$\frac {x^2-2x}{x^2} \le 0$

$\frac {x(x-2)}{x^2} \le 0 / . x^2 > 0 - \forall x$

$x(x-2) \le 0 \Leftrightarrow x \in [0,2]$, but
Pero $x \ne 0$
$\Rightarrow$
$x \in (0,2]$

Respuesta: $x \in (0,2]$

Solución:
$9^{x}-3^{x+1}+2 > 0$

${(3^{2})}^{x}-3^{x}\cdot 3+2 > 0$

${(3^{x})}^{2}-3^{x}\cdot 3+2 > 0$

Sea $3^{x} = u$ , $u>0$

$u^2-3u+2 > 0$
$D=1$
$u_{1}=1$
$u_{2}=2$

$u^2-3u+2 > 0 \Leftrightarrow u \in (-\infty,1) \cup (2,+\infty)$, pero
$u>0 \Rightarrow u \in (0,1) \cup (2,+\infty)$

$\left [ \begin{array}{l} \begin{array}{|l} u > 0 \\ u < 1 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} u > 2 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} 3^{x} > 0 \\ 3^{x} < 3^{0} \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} 3^{x} > 2 / log \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} \forall x \\ x < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} \log_{3}{3^{x}} > log_{3}{2} \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x < 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} x > \log_{3}{2} \end{array} \end{array} \right. \Leftrightarrow$

$x \in (-\infty,0) \cup (\log_{3}{2},+\infty)$

Respuesta: $x \in (-\infty,0) \cup (\log_{3}{2},+\infty)$

Solución:
$4^{x-1} \le 2^{x-2}+3$

$\frac {4^{x}}{4} \le \frac {2^{x}}{2^2} + 3$

$\frac {4^{x}}{4} \le \frac {2^{x}}{4} + 3\ \ \ \ / .4$

$4^{x} \le 2^{x} + 12$
$4^{x} - 2^{x} - 12 \le 0$
${(2^2)}^{x} - 2^{x} - 12 \le 0$
${(2^x)}^{2} - 2^{x} - 12 \le 0$

Sea $2^x = u$ , $u>0$
$u^2-u-12\le 0$
$D=49$
$u_{1}=4$
$u_{2}=-3$

$u^2-u-12\le 0 \Leftrightarrow (u-4)(u+3) \le 0 \Leftrightarrow u \in [-3,4]$, pero
$u>0$
$\Rightarrow u \in (0,4) \Leftrightarrow$

$\begin{array}{|l} u > 0 \\ u \le 4 \end{array} \Leftrightarrow \begin{array}{|l} 2^{x} > 0 \\ 2^{x} \le 4 \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\ 2^{x} \le 2^2 \end{array} \Leftrightarrow x \le 2$

Respuesta: $x \le 2$

Solución:
$2^{x+1} + 2^{2-x} > 9$

$2^{x} \times 2^{1} + \frac {2^2}{2^{x}} > 9$ ; $2^{x} \ne 0$ ; $3a$ $\forall x$

Sea $2^{x}=u$ ; $u>0$

$2u + \frac {4}{u} > 9 / .u$

$2u^2 + 4 > 9u$
$2u^2 -9u + 4 > 0$
$2u^2 -9u + 4 = 0$
$D=49$
$u_{1}=4 > 0$
$u_{2}=\frac {1}{2} > 0$

$\Rightarrow 2u^2 -9u + 4 > 0 \Leftrightarrow 2(u-4)(u-\frac {1}{2}) > 0 /:2$

$(u-4)(u-\frac {1}{2}) > 0 \Leftrightarrow u \in (-\infty,\frac {1}{2}) \cup (4,+\infty)$

pero

$u>0 \Rightarrow u \in (0,\frac {1}{2}) \cup (4,+\infty) \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} u > 0 \\ u < \frac {1}{2} \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} u > 4 \end{array} \end{array} \right. \Leftrightarrow$ $\left [ \begin{array}{l} \begin{array}{|l} 2^{x} > 0 \\ 2^{x} < \frac {1}{2} \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} 2^{x} > 4 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{|l} 2^{x} > 0 \\ 2^{x} < 2^{-1} \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} 2^{x} > 2^2 \end{array} \end{array} \right. \Leftrightarrow$ $\left [ \begin{array}{l} \begin{array}{|l} \forall x \\ x < -1 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} x > 2 \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x < -1 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} x > 2 \end{array} \end{array} \right. \Leftrightarrow$ $x \in (-\infty,-1) \cup (2,+\infty)$

Respuesta: $x \in (-\infty,-1) \cup (2,+\infty)$

Solución:
$2^{x} + 2^{1-x} < 3$

$2^{x} + \frac {2^{1}}{2^{x}} < 3 \qquad /\cdot2^{x} > 0 - \forall x$

$2^{x}\cdot2^{x} + 2 < 3\cdot2^{x}$
$(2^{x})^{2} - 3\cdot2^{x} + 2 < 0$

Sea $2^{x} = u$ ; $u>0$
$u^2 - 3u + 2 < 0$

$D=1$
$u_{1}=1$
$u_{2}=2$
$u^2 - 3u + 2 < 0 \Leftrightarrow x \in (1,2)$

$\left [ \begin{array}{l} \begin{array}{l} u > 1 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} u < 2 \end{array} \end{array} \right. \Leftrightarrow$ $\left [ \begin{array}{l} \begin{array}{l} 2^{x} > 2^{0} \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} 2^{x} < 2^{1} \end{array} \end{array} \right. \Leftrightarrow$ $\left [ \begin{array}{l} \begin{array}{l} x > 0 \end{array} \\ \begin{array}{} \end{array} \\ \begin{array}{l} x < 1 \end{array} \end{array} \right. \Leftrightarrow$ $x \in (0,1)$

Respuesta: $x \in (0,1)$

Solución:
$6^{2x+3} < 6^{x+7} \cdot 3^{2x-8}$

$6^{2x}\cdot6^{3} < 6^{x}\cdot6^{7} \cdot \frac {3^{2x}}{3^{8}} \qquad /\cdot3^{8}$

$3^{8}\cdot6^{2x}\cdot6^{3} < 6^{x}\cdot6^{7}\cdot3^{2x} \qquad / :6^{3}$

$3^{8}\cdot6^{2x} < 6^{x}\cdot6^{4}\cdot3^{2x}$

$(3^{4})^{2}\cdot6^{2x} < 6^{x}\cdot(3\cdot2)^{4}\cdot3^{2x}$

$(3^{4})^{2}\cdot6^{2x} < 6^{x}\cdot3^{4}\cdot2^{4}\cdot3^{2x} \qquad /:{3}^{4} $

$3^{4}\cdot6^{2x} < 6^{x}\cdot2^{4}\cdot3^{2x}$

$3^{4}\cdot(6^{x})^2 < 6^{x}\cdot{2}^{4}\cdot3^{2x} \qquad / :6^{x} \ne 0 - \forall x$

$3^{4}\cdot6^{x} < {2}^{4}\cdot3^{2x} \qquad /:{2}^{4}\cdot6^{x}$

$\frac {3^{4}\cdot6^{x}}{2^{4} \cdot 6^{x}} < \frac {2^{4}\cdot(3^{x})^2}{2^{4}\cdot6^{x}}$

$\frac {3^{4}}{2^{4}} < \frac {(3^{x})^2}{(3\cdot2)^{x}}$

$\frac {3^{4}}{2^{4}} < \frac {(3^{x})^2}{3^{x}\cdot2^{x}}$

$\frac {3^{4}}{{2}^{4}} < \frac {3^{x}}{2^{x}}$

${\left ( \frac {3}{2} \right)}^{4} < {\left ( \frac {3}{2} \right)}^{x}$

$4 < x$
$x > 4$

Respuesta: $x>4$

Solución:
$2^{2x} - 2^{2+2\sqrt{x}} \le 3\cdot 2^{x+\sqrt{x}}$

Dominio: $x\ge 0$

$(2^{x})^2 -2^2\cdot 2^{2\sqrt{x}} \le 3\cdot 2^{x}\cdot 2^{\sqrt{x}}$

$(2^{x})^2 -2^2\cdot (2^{\sqrt{x}})^2 \le 3\cdot 2^{x}\cdot 2^{\sqrt{x}}$

Sea $2^{x}=u$; $u > 0$; $2^{\sqrt{x}} = v$; $v > 0$

$u^2-2^2\cdot v^2 \le 3\cdot u\cdot v$

$u^2-3\cdot u\cdot v-4\cdot v^2 \le 0 \qquad / : v^2$

$\frac {u^2}{v^2}-\frac {3\cdot u\cdot v}{v^2}-\frac {4\cdot v^2}{v^2} \le 0$

${\left (\frac {u}{v} \right )}^{2}-3\frac {u}{v}-4 \le 0$

Sea $\frac {u}{v} = m$; $u>0$; $v>0 \Rightarrow m > 0$

$m^2-3m-4 \le 0$; $m^2-3m-4 = 0$

$D=25$; $m_{1}=4$; $m_{2}=-1$

$m^2-3m-4 \le 0 \Leftrightarrow (m-4)(m+1) \le 0 \Leftrightarrow m \in [-1,4]$

Pero $m > 0 \Rightarrow m \in (0,4]$
$\Leftrightarrow$

$\begin{array}{|l} m > 0 \\\\ m \le 4 \\\\ \end{array} \Leftrightarrow \begin{array}{|l} \frac {u}{v} > 0 \\\\ \frac {u}{v} \le 4 \\ \end{array} \Leftrightarrow \begin{array}{|l} \frac {2^{x}}{2^{\sqrt{x}}} > 0 \\\\ \frac {2^{x}}{2^{\sqrt{x}}} \le 4 \\ \end{array} \Leftrightarrow \begin{array}{|l} 2^{x-\sqrt{x}} > 0 \\\\ 2^{x-\sqrt{x}} \le 2^2 \\ \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\\\ x-\sqrt{x} \le 2 \\ \end{array}$

$x-\sqrt{x} \le 2$
$x-\sqrt{x} -2 \le 0$
$(\sqrt{x})^2-\sqrt{x} -2 \le 0$

Sea $\sqrt{x}=y$ ; $y>0$
$y^2-y-2 \le 0$
$D=9$
$y_{1}=2$
$y_{2}=-1$
$y^2-y-2 \le 0 \Leftrightarrow (y+1)(y-2) \le 0 \Leftrightarrow y \in [-1,2]$

pero
$y>0 \Rightarrow y \in (0,2]$
$\Leftrightarrow$

$\begin{array}{|l} y > 0 \\ y \le 2 \\ \end{array} \Leftrightarrow \begin{array}{|l} \sqrt {x} \ge 0 \\ \sqrt {x} \le 2 \\ \end{array} \Leftrightarrow$ $\begin{array}{|l} \forall x \\ \begin{array}{|l} -4 \le x \le 4 \\ x > 0 \end{array} \end{array} \Leftrightarrow$ $x \in (0,4]$

Respuesta: $x \in (0,4]$

Solución:
$\left | 2^{4x^2-1} -5 \right | \le 3$


$\begin{array}{|l} 2^{4x^2-1} -5 \le 3 \\\\ 2^{4x^2-1} -5 \ge -3 \end{array} \Leftrightarrow \begin{array}{|l} 2^{4x^2-1} \le 8 \\\\ 2^{4x^2-1} \ge 2 \end{array} \Leftrightarrow$

$\begin{array}{|l} 2^{4x^2-1} \le 2^{3} \\\\ 2^{4x^2-1} \ge 2^{1} \end{array} \Leftrightarrow \begin{array}{|l} 4x^2-1 \le 3 \\\\ 4x^2-1 \ge 1 \end{array} \Leftrightarrow \begin{array}{|l} 4x^2 \le 4 \\\\ 4x^2 \ge 2 \end{array} \Leftrightarrow$

$\begin{array}{|l} x^2 \le 1 \\\\ 2x^2 \ge 1 \end{array} \Leftrightarrow \begin{array}{|l} x^2-1 \le 0 \\\\ 2x^2-1 \ge 0 \end{array} \Leftrightarrow$


$\begin{array}{|l} (x-1)(x+1) \le 0 \\\\ (\sqrt{2}x-1)(\sqrt{2}x+1) \ge 0 \end{array} \Leftrightarrow \begin{array}{|l} x \in [-1,1] \\\\ x \in (-\infty,-\frac {\sqrt{2}}{2}) \cup (\frac {\sqrt{2}}{2},+\infty) \end{array} \Leftrightarrow x \in [-1,-\frac{\sqrt{2}}{2}) \cup (\frac{\sqrt{2}}{2},1]$

Respuesta: $x \in [-1,-\frac{\sqrt{2}}{2}) \cup (\frac{\sqrt{2}}{2},1]$

Solución:
$3^{\left | x+2 \right |} + 3^{\left | x-1 \right |} \ge 28$

$\begin{array}{c} \\\\ x+2 \\\\ x-1 \end{array} \begin{array}{c} (-\infty,-2) \\\\ - \\\\ - \end{array} \begin{array}{c} [-2,1) \\\\ + \\\\ - \end{array} \begin{array}{c} [1,+\infty) \\\\ + \\\\ + \end{array}$

---

Caso I. $x \in (-\infty,-2)$

$3^{-(x+2)} + 3^{-(x-1)} \ge 28$

$\frac {1}{3^{(x+2)}} + 3^{1-x} \ge 28$

$\frac {1}{3^{x}\cdot3^{2}} + \frac {3^{1}}{3^{x}} \ge 28$

Let $3^{x}=u$; $u>0$

$\frac {1}{u\cdot9} + \frac {3}{u} \ge 28$

$\frac {1}{u\cdot9} + \frac {3}{u} - 28 \ge 0$

$\frac {1+27-28\cdot9u}{9u} \ge 0 \qquad/ \cdot 9u$

$1+27-252u \ge 0$
$252u \le 28 \qquad / : 28$
$9u \le 1$
$u \le \frac {1}{9}$; pero
$u>0 \Leftrightarrow u \in (0,\frac {1}{9}]$

$\begin{array}{|l} u>0 \\\\ u \le \frac {1}{9} \end{array} \Leftrightarrow \begin{array}{|l} 3^{x}>0 \\\\ 3^{x} \le \frac {1}{9} \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\\\ 3^{x} \le 3^{-2} \end{array} \Leftrightarrow x \le -2$

pero
Case I. $x \in (-\infty,-2)$
$\Rightarrow x \in (-\infty,-2)$

---

Caso II. $x \in [-2,1)$

$3^{(x+2)} + 3^{-(x-1)} \ge 28$

$3^{(x+2)} + 3^{1-x} \ge 28$

$3^{x}\cdot 3^{2} + \frac {3^{1}}{3^{x}} \ge 28$

Sea $3^{x}=u$; $u>0$

$9u + \frac {3}{u} \ge 28$

$9u + \frac {3}{u} - 28 \ge 0$

$\frac {9u^2 + 3 - 28u}{u} \ge 0 \qquad / . u$

$9u^2-28u+3 \ge 0$

$D=676$

$u_{1}=3 > 0$; $u_{2}=\frac {1}{9} > 0$

$9u^2-28u+3 \ge 0 \Rightarrow 9(u-3)(u-\frac {1}{9}) \ge 0 \Leftrightarrow$

$u \in (-\infty,\frac {1}{9}] \cup [3,+\infty)$; pero

$u>0 \Rightarrow u \in (0,\frac {1}{9}] \cup [3,+\infty) \Leftrightarrow$

$\begin{array}{|l} u>0 \\\\ u \le \frac {1}{9} \end{array} \Leftrightarrow \begin{array}{|l} 3^{x}>0 \\\\ 3^{x} \le \frac {1}{9} \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\\\ 3^{x} \le 3^{-2} \end{array} \Leftrightarrow x \le -2$

o

$u \ge 3 \Leftrightarrow 3^{x} \ge 3^{1} \Leftrightarrow x \ge 1$
$x \in (-\infty,-2] \cup [1,+\infty)$, pero
II. $x \in [-2,1)$
$\Rightarrow x = -2$

---

Caso III. $x \in [1,+\infty)$

$3^{(x+2)} + 3^{(x-1)} \ge 28$

$3^{x}\cdot 3^{2} + 3^{x}\cdot 3^{-1} \ge 28$

$3^{x}\cdot 9 + \frac {3^{x}}{3} \ge 28 \qquad / \cdot 3$

$3\cdot 3^{x}\cdot 9 + 3^{x} \ge 84$

$28\cdot 3^{x} \ge 84$

$3^{x} \ge 3$

$3^{x} \ge 3^{1} \Rightarrow x \ge 1$

Caso III. $x \in [1,+\infty)$
$\Rightarrow x \in [1,+\infty)$

Respuesta: $x \in (-\infty,-2] \cup [1,+\infty) $

Solución:
$3^{\left | x+3 \right |} - \left | 3^{x+2} -1 \right | < 3^{x+2} + 2$

$x+3=0 \Rightarrow x=-3$

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


$\begin{array}{c} \\\\ x+3 \\\\ 3^{x+2}-1 \end{array} \begin{array}{c} (-\infty,-3) \\\\ - \\\\ - \end{array} \begin{array}{c} [-3,-2) \\\\ + \\\\ - \end{array} \begin{array}{c} [-2,+\infty) \\\\ + \\\\ + \end{array}$


I. $x \in (-\infty,-3)$

$3^{-(x+3)} -(-(3^{x+2}-1)) < 3^{x+2}+2$

$3^{-(x+3)} +(3^{x+2}-1) < 3^{x+2}+2$

$3^{-(x+3)} +3^{x+2}-1 < 3^{x+2}+2$

$3^{-(x+3)} < 3$

$3^{-(x+3)} < 3^{1}$

$-(x+3) < 1$
$-x-3 < 1$
$x > -4$ peroa
$x \in (-\infty,-3) \Rightarrow x \in (-4,-3)$

---

II. $x \in [-3,-2)$

$3^{(x+3)} -(-(3^{x+2}-1)) < 3^{x+2}+2$

$3^{(x+3)} +(3^{x+2}-1) < 3^{x+2}+2$

$3^{(x+3)} +3^{x+2}-1 < 3^{x+2}+2$

$3^{(x+3)} < 3$

$3^{(x+3)} < 3^{1}$

$x+3 < 1$
$x < -2$; pero
$x \in [-3,-2) \Rightarrow x \in [-3,-2)$

---

III. $x \in [-2,+\infty)$

$3^{(x+3)} -(3^{x+2}-1) < 3^{x+2}+2$

$3^{(x+3)} -3^{x+2} +1 < 3^{x+2}+2$

$3^{x}\cdot 3^{3} -3^{x}\cdot 3^{2} +1 < 3^{x}\cdot 3^{2}+2$

$27\cdot 3^{x} -9\cdot 3^{x} +1 - 9 \cdot 3^{x} -2 < 0$

$9 \cdot 3^{x} < 1$

$3^{x} < \frac {1}{9}$

$3^{x} < 9^{-1}$

$3^{x} < 3^{-2}$

$x<-2$; pero

$x \in [-2,+\infty) \Rightarrow x \in \empty$

Respuesta: $x \in (-4,-2)$

Solución:
$4^{2x} - 2\cdot 4^{x} < 0$

$(4^{x})^2 - 2\cdot 4^{x} < 0$

Sea $4^{x}=u$; $u>0$

$u^2-2u<0$

$u(u-2)<0 \Leftrightarrow u \in (0,2) \Leftrightarrow \begin{array}{|l} u>0 \\\\ u<2 \end{array} \Leftrightarrow$

$\begin{array}{|l} 4^{x}>0 \\\\ 4^{x}<2 \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\\\ (2^2)^{x}<2 \end{array} \Leftrightarrow$

$\begin{array}{|l} \forall x \\\\ 2^{2x}<2^{1} \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\\\ 2x<1 \end{array} \Leftrightarrow \begin{array}{|l} \forall x \\\\ x<\frac {1}{2} \end{array} \Leftrightarrow$

$\Leftrightarrow x<\frac {1}{2} \Leftrightarrow x \in (-\infty,\frac{1}{2})$

Respuesta: $x \in (-\infty,\frac{1}{2})$

Solución:
$3\cdot 3^{2x} + 11 \cdot 3^{x} -4 > 0$

$3\cdot (3^{x})^2 + 11 \cdot 3^{x} -4 > 0$

Sea $3^{x}=u$; $u>0$

$3u^2 + 11u -4 > 0$

$3u^2 + 11u -4 = 0$

$D=169$
$u_{1}=-4$
$u_{2}=\frac {1}{3}$

$3u^2 + 11u -4 > 0 \Leftrightarrow (u+4)(u-\frac {1}{3}) > 0 \Leftrightarrow u \in (-\infty,-4) \cup (\frac {1}{3},+\infty) \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} u < -4 \\\\ u > \frac {1}{3} \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} 3^{x} < -4 \\\\ 3^{x} > \frac {1}{3} \end{array} \end{array} \right. \Leftrightarrow$

$\left [ \begin{array}{l} \begin{array}{l} x \in \empty \\\\ 3^{x} > 3^{-1} \end{array} \end{array} \right. \Leftrightarrow$

$\Leftrightarrow \left [ \begin{array}{l} \begin{array}{l} x \in \empty \\\\ x > -1 \end{array} \end{array} \right. \Leftrightarrow$

$x>-1 \Leftrightarrow x \in (-1,+\infty)$

Respuesta: $x \in (-1,+\infty)$

Solución:
[tex]-2\lt -0{,} 6[/tex]
La potencia con el exponente menor es la mayor, así que la función [tex]t\mapsto a^t[/tex] es decreciente, es decir, [tex]0\lt a\lt 1[/tex]. Para [tex]a=1[/tex] ambos miembros valen [tex]1[/tex]; para [tex]a\gt 1[/tex] la función es creciente y [tex]a^{-2}\lt a^{-0{,} 6}[/tex].

Solución:
[tex]3^x-5^x\ge 0[/tex]. Dividimos entre [tex]5^x\gt 0[/tex]: [tex]\left(\dfrac{3}{5}\right)^x\ge 1=\left(\dfrac{3}{5}\right)^0[/tex]
La base es menor que [tex]1[/tex], así que el sentido de la desigualdad se invierte: [tex]x\le 0[/tex]

Solución:
[tex]0{,} 4=\dfrac{2}{5}; \dfrac{125}{8}=\left(\dfrac{5}{2}\right)^3=\left(\dfrac{2}{5}\right)^{-3}[/tex]
[tex]\left(\dfrac{2}{5}\right)^x\ge \left(\dfrac{2}{5}\right)^{-3}[/tex]. La base es menor que [tex]1[/tex], así que el sentido de la desigualdad se invierte: [tex]x\le -3[/tex]

Solución:
[tex]3=2^{\log_2 3}; 5=2^{\log_2 5}[/tex]
[tex]2^x[/tex] es estrictamente creciente, por lo tanto [tex]\log_2 3\lt x\lt \log_2 5[/tex].

Solución:
[tex]6^x(1+6)=7\times 6^x; 2^x(1+2+4)=7\times 2^x[/tex]
[tex]6^x\lt 2^x \iff 3^x\lt 1 \iff x\lt 0[/tex]

Solución:
[tex]3^{2x}\left(6+\dfrac{1}{3}-5\right)=\dfrac{4}{3}\times 3^{2x}\le 36 \iff 3^{2x}\le 27=3^3 \iff x\le \dfrac{3}{2}[/tex]
Respuesta: [tex]1[/tex]

Solución:
[tex]f(x)=3^x+x-1[/tex] es estrictamente creciente (suma de funciones crecientes); [tex]f(0)=0[/tex].
[tex]f(x)\gt 0=f(0) \iff x\gt 0[/tex]