Question

Me ajudem por favor

Answer 1

1) $2^{\frac{45}{x}}=32$

$2^{\frac{45}{x}}=2^{5}$

$\dfrac{45}{x}=5$

$5x=45$

$x=\dfrac{45}{5}$

$\boxed{x=9}$

Letra D


2) $\left(\dfrac{1}{5}\right)^{2x-1}=125^{x+5}$

$(5^{-1})^{2x-1}=(5^3)^{x+5}$

$5^{-2x+1}=5^{3x+15}$

$-2x+1=3x+15$

$-5x=14$

$\boxed{x=-\dfrac{14}{5}}$

Letra E


3) $4^{2-3x}=\sqrt{2}$

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

$2^{4-6x}=2^{\frac{1}{2}}$

$4-6x=\dfrac{1}{2}$

$8-12x=1$

$12x=7$

$\boxed{x=\dfrac{7}{12}}$

$\dfrac{7}{12}\in[0,1]$, pois $0<\dfrac{7}{12}<1$

Letra B


4) $4^{x^2-8x+15}=1$

$4^{x^2-8x+15}=4^{0}$

$x^2-8x+15=0$

$\Delta=(-8)^2-4\cdot1\cdot15=64-60=4$

$x=\dfrac{-(-8)\pm\sqrt{4}}{2\cdot1}=\dfrac{8\pm2}{2}=4\pm1$

$x'=4+1 \iff \boxed{x'=5}$

$x"=4-1 \iff \boxed{x"=3}$

Letra C


5) $2^{3x+2}=32$

$2^{3x+2}=2^{5}$

$3x+2=5$

$3x=3$

$\boxed{x=1}$

Letra A


6) $P(t)=18\cdot5^{t-5}$

$18\cdot5^{t-5}=2250$

$5^{t-5}=\dfrac{2250}{18}$

$5^{t-5}=125$

$5^{t-5}=5^3$

$t-5=3$

$\boxed{t=8}$

Letra B


7) $2^{x+1}-2^{3-x}=6$

$2\cdot2^{x}-\dfrac{2^{3}}{2^{x}}=6$

$2^{x}=y$

$2y-\dfrac{8}{y}=6$

$2y^2-6y-8=0$

$y^2-3y-4=0$

$\Delta=(-3)^2-4\cdot1\cdot(-4)=9+16=25$

$y=\dfrac{-(-3)\pm\sqrt{25}}{2\cdot1}=\dfrac{3\pm5}{2}$

$y'=\dfrac{3+5}{2}=\dfrac{8}{2}=4$

$y"=\dfrac{3-5}{2}=\dfrac{-2}{2}=-1$ (não serve, pois $2^{x}>0$)

$2^{x}=4$

$2^{x}=2^2$

$x=2$

$x^2+20=2^2+20=4+20 \iff \boxed{x^2+20=24}$

Letra B


8) $5^{x}+5^{x-1}+5^{x-2}+5^{x-3}=156$

$5^{x-3}\cdot5^{3}+5^{x-3}\cdot5^{2}+5^{x-3}\cdot5^{1}+5^{x-3}=156$

$5^{x-3}\cdot(5^3+5^2+5^1+1)=156$

$5^{x-3}\cdot156=156$

$5^{x-3}=\dfrac{156}{156}$

$5^{x-3}=1$

$5^{x-3}=5^{0}$

$x-3=0$

$\boxed{x=3}$

Letra E


9) $3^{2x}-10\cdot3^{x}+9=0$

$(3^{x})^2-10\cdot3^{x}+9=0$

$3^{x}=y$

$y^2-10y+9=0$

$\Delta=(-10)^2-4\cdot1\cdot9=100-36=64$

$y=\dfrac{-(-10)\pm\sqrt{64}}{2\cdot1}=\dfrac{10\pm8}{2}=5\pm4$

$y'=5+4=9$

$y"=5-4=1$

Para $y=9$:

$3^{x}=9$

$3^{x}=3^2$

$x'=2$

Para $y=1$:

$3^x=1$

$3^x=3^0$

$x"=0$

O produto é $x'\cdot x"=2\cdot0 \iff \boxed{x'\cdot x"=0}$

Letra A


10) Uma função exponencial $f(x)=a^{x}$ é crescente se $a>0$ e decrescente se $0<a<1$

a) $f(x)=2^{x}$ é crescente, pois $2>0$

$f(-1)=2^{-1}=\dfrac{1}{2^1}=\dfrac{1}{2}$

$f(0)=2^{0}=1$

$f(1)=2^1=2$

$f(2)=2^2=4$

$f(3)=2^3=8$

b) $f(x)=\left(\dfrac{1}{3}\right)^{x}$ é decrescente, pois $0<\dfrac{1}{3}<1$

$f(-2)=\left(\dfrac{1}{3}\right)^{-2}=3^2=9$

$f(-1)=\left(\dfrac{1}{3}\right)^{-1}=3^1=3$

$f(0)=\left(\dfrac{1}{3}\right)^{0}=1$

$f(1)=\left(\dfrac{1}{3}\right)^{1}=\dfrac{1}{3}$

Os gráficos estão na imagem em anexo