Question

valendo 60 pontos para quem resolver essas contas ate amanha ao meio dia. PFVR me ajudem
Para quem teve dificuldade em ver a foto as contas sao essas.
a)2^x-9=24/6

b)7^x+1 -49.7^x-2=42

c)-6.5^x +25^x=-5

d)3^x+1 +3^x=36

e)9^x +3.3^x-1=12

f)16^x-48/16=2^2x-3

Answer 1

a) $2^{x-9}=\dfrac{24}{6}$

$2^{x-9}=4$

$2^{x-9}=2^2$

$x-9=2$

$x=11$


b) $7^{x+1}-49\cdot7^{x-2}=42$

$7\cdot7^{x}-7^2\cdot7^{x-2}=42$

$7\cdot7^{x}-7^{x}=42$

$6\cdot7^{x}=42$

$7^{x}=7$

$7^{x}=7^{1}$

$x=1$


c) $-6\cdot5^{x}+25^{x}=-5$

$-6\cdot5^{x}+(5^2)^{x}=-5$

$-6\cdot5^{x}+5^{2x}=-5$

$-6\cdot5^{x}+5^{x}\cdot5^{x}=-5$

$5^{x}=y$

$-6y+y^2=-5$

$y^2-6y+5=0$

$\Delta=(-6)^2-4\cdot1\cdot5=36-20=16$

$y=\dfrac{-(-6)\pm\sqrt{16}}{2}=\dfrac{6\pm4}{2}=3\pm2$

$y=3+2=5$ ou $y=3-2=1$

$5^{x}=5$

$x=1$

ou $5^{x}=1$

$x=0$


d) $3^{x+1}+3^{x}=36$

$3\cdot3^{x}+3^{x}=36$

$3^{x}=y$

$3y+y=36$

$4y=36$

$y=9$

$3^{x}=9$

$x=2$.


e) $9x^{x}+3\cdot3^{x-1}=12$

$(3^2)^{x}+3^{x}=12$

$3^{x}\cdot3^{x}+3^{x}=12$

$y=3^{x}$

$y^2+y=12$

$y(y+1)=12$

$y=3$ ou $y=-4$ (não serve)

$3^x=3$

$x=1$


f) $\dfrac{16^{x}-48}{16}=2^{2x-3}$

$\dfrac{(2^4)^{x}-48}{2^4}=2^{2x-3}$

$2^{4x}-48=2^{2x-3+4}$

$2^{4x}-48=2^{2x+1}$

$2^{x}\cdot2^{x}\cdot2^{x}\cdot2^{x}-48=2^{x}\cdot2^{x}\cdot2$

$y=2^{x}$

$y^4-48=2y^2$

$y^4-2y^2-48=0$

$k=y^2$

$k^2-2k-48=0$

$\Delta=(-2)^2-4\cdot1\cdot(-48)=4+192=196$

$k=\dfrac{-(-2)\pm\sqrt{196}}{2}=\dfrac{2\pm14}{2}=1\pm7$

$k=1+7=8$ ou $k=1-7=-6$ (não serve)

$y^2=8$

$y=2\sqrt{2}$

$2^{x}=2\sqrt{2}$

$2^{x}=\sqrt{8}$

$2^{x}=\sqrt{2^3}$

$2^{x}=2^{\frac{3}{2}}$

$x=\dfrac{3}{2}$

Answer 2

$a) \ 2^{x+1}=\frac{24}{6}\\ \\ 2^{x+1}=2^2\\ \\ x+1=2\\ \\ \boxed{x=1}\\ \\ b) \ 7^{x+1}-49.7^{x-2}=42\\ \\ 7.7^x-7^x=42\\ \\ y=7^x\\ \\ 7y-y=42\\ \\ 6y=42\\ \\ y=7\\ \\ 7^x=7^1\\ \\ \boxed{x=1}$

$c) \ -6.5^x+25^x=-5\\ \\ -6.5^x+(5^x)^2=-5\\ \\ y=5^x\\ \\ -6y+y^2+5-0\\ \\ y=1 \ \ ou \ \ \ y=5\\ \\ 5^x=5\Rightarrow x=1\\ \\ 5^x=1\Rightarrow x =0$

$d) \ 3^{x+1}+3^x=36\\ \\ 3.3^x+3^x=36\\ \\ 4.3^x=36\\ \\ 3^x=9\\ \\ 3^x=3^2\\ \\ \boxed{x=2}$

$e) \ 9^x+3.3^{x-1}=12\\ \\ (3^x)^2+3^x=12\\ \\ y=3^x\\ \\ y^2+y-12=0\\ \\ y=-4 \ \ \ ou \ \ \ y=3\\ \\ 3^x=3\\ \\ \boxed{x=3}$\\
\\
$f) \ \frac{16^x-48}{16}=2^{2x-3}\\ \\ \frac{16^x}{16}-\frac{48}{16}=\frac{4^x}{8}\\ \\ 16^x-48=2.4^x\\ \\ (4^x)^2-2.4^x-48=0\\ \\ y=4^x\\ \\ y^2-2y-48=0\\ \\ y=8\\ \\ 4^x=8\\ \\ 2^{2x}=2^3\\ \\ 2x=3\\ \\ x=\frac{3}{2}$