Question

15).Resolva as seguintes equações exponenciais.

Answer 1

a) $2^{4x+1}\cdot8^{-x+3}=\dfrac{1}{16}$

$2^{4x+1}\cdot(2^3)^{-x+3}=\dfrac{1}{2^4}~\longrightarrow~2^{4x+1}\cdot2^{-3x+9}=2^{-4}$

$2^{4x+1-3x+9}=2^{-4}~\longrightarrow~2^{x+10}=2^{-4}$

Igualando os expoentes:

$x+10=-4~\longrightarrow~x=-4-10~\longrightarrow~\boxed{x=-14}$

$\text{S}=\{-14\}$


b) $\dfrac{\left(\dfrac{1}{5}\right)^{3x}}{25^{2+x}}=5$

$\dfrac{(5^{-1})^{3x}}{(5^2)^{2+x}}=5^1~\longrightarrow~\dfrac{5^{-3x}}{5^{4+2x}}=5^1$

$5^{-3x-4-2x}=5^1~\longrightarrow~5^{-5x-4}=5^{1}$

Igualando os expoentes:

$-5x-4=1~\longrightarrow~-5x=1+4~\longrightarrow~-5x=5$

$x=\dfrac{5}{-5}~\longrightarrow~\boxed{x=-1}$

$\text{S}=\{-1\}$


c) $\left(\dfrac{1}{9}\right)^{x^2-1}\cdot27^{1-x}=3^{2x+7}$

$(3^{-2})^{x^2-1}\cdot(3^3)^{1-x}=3^{2x+7}~\longrightarrow~3^{-2x^2+2}\cdot3^{3-3x}=3^{2x+7}$

$3^{-2x^2+2+3-3x}=3^{2x+7}~\longrightarrow~3^{-2x^2-3x+5}=3^{2x+7}$

Igualando os expoentes:


$-2x^2-3x+5=2x+7~\longrightarrow~-2x^2-3x-2x+5-7=0$

$-2x^2-5x-2=0~\longrightarrow~2x^2+5x+2=0$

$\Delta=5^2-4\cdot2\cdot2=25-16=9$

$x=\dfrac{-5\pm\sqrt{9}}{2\cdot2}=\dfrac{-5\pm3}{4}$

$x'=\dfrac{-5+3}{4}~\longrightarrow~x'=\dfrac{-2}{4}~\longrightarrow~\boxed{x'=-\dfrac{1}{2}}$

$x"=\dfrac{-5-3}{4}~\longrightarrow~x"=\dfrac{-8}{4}~\longrightarrow~\boxed{x"=-2}$

$\text{S}=\{-2,-\frac{1}{2}\}$



d) $(\sqrt{10})^{x}\cdot(0,01)^{4x-1}=\dfrac{1}{1000}$

$(10^{\frac{1}{2}})^{x}\cdot(10^{-2})^{4x-1}=\dfrac{1}{10^3}~\longrightarrow~10^{\frac{x}{2}}\cdot10^{-8x+2}=10^{-3}$

$10^{\frac{x}{2}-8x+2}=10^{-3}~\longrightarrow~10^{\frac{x-16x+4}{2}}=10^{-3}~\longrightarrow~10^{\frac{-15x+4}{2}}=10^{-3}$

Igualando os expoentes:

$\dfrac{-15x+4}{2}=-3$

$-15x+4=2\cdot(-3)~\longrightarrow~-15x+4=-6~\longrightarrow~-15x=-6-4$

$-15x=-10~\longrightarrow~x=\dfrac{-10}{-15}~\longrightarrow~\boxed{x=\dfrac{2}{3}}$

$\text{S}=\{\frac{2}{3}\}$