Question

função exponencial..........​

Answer 1

Explicação passo-a-passo:

14)

a)

$\sf 9^{x}\cdot\sqrt{3}=1$

$\sf (3^2)^{x}\cdot3^{\frac{1}{2}}=3^0$

$\sf 3^{2x}\cdot3^{\frac{1}{2}}=3^0$

$\sf 3^{2x+\frac{1}{2}}=3^0$

Igualando os expoentes:

$\sf 2x+\dfrac{1}{2}=0$

$\sf 2\cdot2x+1=0$

$\sf 4x+1=0$

$\sf 4x=-1$

$\sf \red{x=\dfrac{-1}{4}}$

b)

$\sf 5\cdot25^x=\dfrac{5^x}{5}$

$\sf 5\cdot(5^2)^x=\dfrac{5^x}{5}$

$\sf 5\cdot(5^x)^2=\dfrac{5^x}{5}$

Seja $\sf y=5^x$

$\sf 5\cdot y^2=\dfrac{y}{5}$

$\sf 5\cdot 5\cdot y^2=y$

$\sf 25y^2=y$

$\sf 25y^2-y=0$

$\sf y\cdot(25y-1)=0$

=> $\sf y'=0$ (não serve, pois $\sf 5^x > 0)$, para todo x real)

=> $\sf 25y-1=0$

$\sf 25y=1$

$\sf y=\dfrac{1}{25}$

Assim:

$\sf 5^x=\dfrac{1}{25}$

$\sf 5^x=\dfrac{1}{5^2}$

$\sf 5^x=5^{-2}$

Igualando os expoentes:

$\sf \red{x=-2}$

15)

$\sf 3^{x+3}+3^{x+2}+3^{x+1}+3^x+3^{x-1}=121$

$\sf 3^{x-1}\cdot(3^4+3^3+3^2+3^2+3^1+1)=121$

$\sf 3^{x-1}\cdot(81+27+9+3+1)=121$

$\sf 3^{x-1}\cdot121=121$

$\sf 3^{x-1}=\dfrac{121}{121}$

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

$\sf 3^{x-1}=3^0$

Igualando os expoentes:

$\sf x-1=0$

$\sf \red{x=1}$

16)

$\sf 2^{2x+2}-3\cdot2^{x+2}=16$

$\sf 2^{2x}\cdot2^2-3\cdot2^{x}\cdot2^2=16$

$\sf (2^x)^2\cdot4-3\cdot2^{x}\cdot4=16$

$\sf 4\cdot(2^x)^2-12\cdot2^x-16=0$

Seja $\sf y=2^x$

$\sf 4y^2-12y-16=0$

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

$\sf \Delta=(-3)^2-4\cdot1\cdot(-4)$

$\sf \Delta=91+16$

$\sf \Delta=25$

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

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

$\sf y"=\dfrac{3-5}{2}~\Rightarrow~y"=\dfrac{-2}{2}~\Rightarrow~y"=-1$ (não serve)

Assim:

$\sf 2^x=4$

$\sf 2^x=2^2$

Igualando os expoentes:

$\sf \red{x=2}$

17)

$\sf 7^{2x}=8\cdot7^{x}-7$

$\sf 7^{2x}-8\cdot7^{x}+7=0$

$\sf (7^x)^2-8\cdot7^{x}+7=0$

Seja $\sf y=7^x$

$\sf y^2-8y+7=0$

$\sf \Delta=(-8)^2-4\cdot1\cdot7$

$\sf \Delta=64-28$

$\sf \Delta=36$

$\sf y=\dfrac{-(-8)\pm\sqrt{36}}{2\cdot1}=\dfrac{8\pm6}{2}$

$\sf y'=\dfrac{8+6}{2}~\Rightarrow~y'=\dfrac{14}{2}~\Rightarrow~y'=7$

$\sf y"=\dfrac{8-6}{2}~\Rightarrow~y"=\dfrac{2}{2}~\Rightarrow~y"=1$

=> Para y = 7:

$\sf 7^x=7$

$\sf 7^x=7^1$

Igualando os expoentes:

$\sf \red{x=1}$

=> Para y = 1:

$\sf 7^x=1$

$\sf 7^x=7^0$

Igualando os expoentes:

$\sf \red{x=0}$

O conjunto solução é $\sf S=\{0,1\}$

18)

$\sf \begin{cases} \sf 2^{x}+2^{y}=20 \\ \sf 2^{x+y}=64 \end{cases}$

Da segunda equação:

$\sf 2^{x+y}=64$

$\sf 2^{x+y}=2^6$

Igualando os expoentes:

$\sf x+y=6$

$\sf x=6-y$

Substituindo na primeira equação:

$\sf 2^{6-y}+2^{y}=20$

$\sf \dfrac{2^6}{2^y}+2^{y}=20$

$\sf \dfrac{64}{2^y}+2^{y}=20$

Seja $\sf k=2^y$

$\sf \dfrac{64}{k}+k=20$

$\sf 64+k^2=20k$

$\sf k^2-20k+64=0$

$\sf \Delta=(-20)^2-4\cdot1\cdot64$

$\sf \Delta=400-256$

$\sf \Delta=144$

$\sf y=\dfrac{-(-20)\pm\sqrt{144}}{2\cdot1}=\dfrac{20\pm12}{2}$

$\sf k'=\dfrac{20+12}{2}~\Rightarrow~k'=\dfrac{32}{2}~\Rightarrow~k'=16$

$\sf k'=\dfrac{20-12}{2}~\Rightarrow~k'=\dfrac{8}{2}~\Rightarrow~k'=4$

=> Para k = 16:

$\sf 2^y=16$

$\sf 2^y=2^4$

Igualando os expoentes:

$\sf \red{y=4}$

Assim:

$\sf x=6-y$

$\sf x=6-4$

$\sf \red{x=2}$

=> Para k = 4:

$\sf 2^y=4$

$\sf 2^y=2^2$

Igualando os expoentes:

$\sf \red{y=2}$

Assim:

$\sf x=6-y$

$\sf x=6-2$

$\sf \red{x=4}$

O conjunto solução é $\sf S=\{(2,4),(4,2)\}$

19)

Temos que:

$\sf f(x)=5^{3x}$

=> $\sf f(k)=27$

$\sf 5^{3k}=27$

Assim:

$\sf f\Big(\dfrac{k}{6}\Big)=5^{3\cdot\frac{k}{6}}$

$\sf f\Big(\dfrac{k}{6}\Big)=5^{\frac{3k}{6}}$

$\sf f\Big(\dfrac{k}{6}\Big)=5^{\frac{k}{2}}$

$\sf \Big[(f\Big(\dfrac{k}{6}\Big)\Big]^{6}=\Big(5^{\frac{k}{2}}\Big)^{6}$

$\sf \Big[(f\Big(\dfrac{k}{6}\Big)\Big]^{6}=5^{\frac{6k}{2}}$

$\sf \Big[(f\Big(\dfrac{k}{6}\Big)\Big]^{6}=5^{3k}$

Substituindo $\sf 5^{3k}~por~27$:

$\sf \Big[(f\Big(\dfrac{k}{6}\Big)\Big]^{6}=27$

$\sf f\Big(\dfrac{k}{6}\Big)=\sqrt[6]{27}$

$\sf f\Big(\dfrac{k}{6}\Big)=\sqrt[6]{3^3}$

$\sf f\Big(\dfrac{k}{6}\Big)=\sqrt{3}$