Question

Resolva as equações exponenciais.

Answer 1

Olá, Thatta.

$f)\,2^{\frac x 5}\cdot(2^2)^{\frac x 3}=(2^3)^{\frac{-x}{2}}\Rightarrow\\\\ \frac x 5+\frac{2x}3=-\frac{3x}{2}\Rightarrow\\\\ \frac{6x+20x}{30}=-\frac{45x}{30}\Rightarrow\\\\ 26x=-45x\Rightarrow\\\\ \boxed{x=0}$


$g)\,(3^{-1})^{x+1}=\frac{3^{\frac12}}{3^2}\Rightarrow\\\\ 3^{-x-1}=3^{\frac12-2}\Rightarrow\\\\ -x-1=\frac{1-4}2\Rightarrow\\\\ -x = \frac32+1\Rightarrow\\\\ -x=\frac{3+2}2\Rightarrow\\\\ \boxed{x=-\frac52}$

Answer 2

Explicação passo-a-passo:

f)

$\sf \sqrt[5]{2^x}\cdot\sqrt[3]{4^x}=\sqrt{8^{-x}}$

$\sf \sqrt[5]{2^x}\cdot\sqrt[3]{(2^2)^x}=\sqrt{(2^3)^{-x}}$

$\sf \sqrt[5]{2^x}\cdot\sqrt[3]{2^{2x}}=\sqrt{2^{-3x}}$

$\sf 2^{\frac{x}{5}}\cdot2^{\frac{2x}{3}}=2^{\frac{-3x}{2}}$

$\sf 2^{\frac{x}{5}+\frac{2x}{3}}=2^{\frac{-3x}{2}}$

$\sf 2^{\frac{3x+10x}{15}}=2^{\frac{-3x}{2}}$

$\sf 2^{\frac{13x}{15}}=2^{\frac{-3x}{2}}$

Igualando os expoentes:

$\sf \dfrac{13x}{15}=\dfrac{-3x}{2}$

$\sf \dfrac{13x}{15}+\dfrac{3x}{2}=0$

$\sf 26x+45x=0$

$\sf 71x=0$

$\sf x=\dfrac{0}{71}$

$\sf \red{x=0}$

O conjunto solução é $\sf S=\{0\}$

g)

$\sf \Big(\dfrac{1}{3}\Big)^{x+1}=\dfrac{\sqrt{3}}{9}$

$\sf (3^{-1})^{x+1}=\dfrac{3^{\frac{1}{2}}}{3^2}$

$\sf 3^{-x-1}=3^{\frac{1}{2}-2}$

$\sf 3^{-x-1}=3^{\frac{1-4}{2}}$

$\sf 3^{-x-1}=3^{\frac{-3}{2}}$

Igualando os expoentes:

$\sf -x-1=\dfrac{-3}{2}$

$\sf x=-1+\dfrac{3}{2}$

$\sf x=\dfrac{-2+3}{2}$

$\sf \red{x=\dfrac{1}{2}}$

O conjunto solução é $\sf S=\Big\{\dfrac{1}{2}\Big\}$