Question

Resolva as seguintes equações exponenciais e determine o conj. verdade
a) $2^{x-3}$ + $2^{x-1}$ + $2^{x}$ = 52

b) $\frac{25^{x}+125}{6} = 5^{x} + 1$

Answer 1

Ola Green

a)

2^(x - 3) + 2^(x - 1) + 2^x = 52

2^x/8 + 2^x/2 + 2^x = 52

y = 2^x

y/8 + y/2 + y = 52

y + 4y + 8y = 416 

13y = 416

y = 416/13 = 32

2^x = 32 = 2^5

x = 5
S = (5)

b)

(25^x + 125)/6 = 5^x + 1

5^(2x) + 125 = 6*5^x + 6

y = 5^x

y² - 6y + 119 = 0 

delta 
d² = 36 - 4*119

d < 0 

S = (Ø)

Answer 2

$2^{x-3}+2^{x-1}+2^x=52\\ 2^x\cdot2^{-3}+2^x\cdot2^{-1}+2^x=52\\\\ evidencia~2^x\\\\ 2^x\cdot(2^{-3}+2^{-1}+1)=52\\ 2^x\cdot\left( \dfrac{1}{2^3}+ \dfrac{1}{2^1}+1\right)=52\\ 2^x\cdot\left( \dfrac{1}{8}+ \dfrac{1}{2}+1\right)=52\\ 2^x\cdot\left( \dfrac{10}{16} +1\right)=52\\ 2^x\cdot\left( 1\dfrac{5}{8}\right)=52\\ 2^x\cdot \dfrac{13}{8}=52\\ 2^x\cdot 13=52\cdot8\\ 2^x\cdot13=416\\ 2^x=416/13\\ 2^x=32\\ 2^x=2^5\\ \not2^x=\not2^5\\\\ x=5\\\\ \huge\boxed{\text{S}=\{5\}}$

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$\dfrac{25^x+125}{6}=5^x+1\\\\ 25^x+125=6\cdot(5^x+1)\\ (5^2)^x+125=6\cdot5^x+6\\ (5^x)^2+125=6\cdot5^x+6\\ (5^x)^2-6\cdot5^x+125-6=0\\ (5^x)^2-6\cdot5^x+119=0\\\\ 5^x=y\\\\ y^2-6y+119=0\\\\ \Delta=(-6)^2-4\cdot1\cdot119\\ \Delta=36-476\\ \Delta=-440 ~~\notin~\mathbb{R}\\\\\\ \huge\boxed{\text{S}=\{\O\}}$