Question

Exercício sobre equação exponencial:

Resolva: 4^(x+1)-9*2^(x)=0
e
3^(x-1)-3^(x)+3^(x+1)+3^(x+2)=306

Answer 1

Resposta:

Explicação passo a passo:

a)

$4^{x+1}-9\cdot2^x+2=0\\4^x\cdot4^1-9\cdot2^x+2=0\\(2^2)^x\cdot 4-9\cdot2^x+2=0\\4(2^x)^2-9\times 2^x+2=0\\2^x=t\\4t^2-9t+2=0\\\Delta = (-9)^2-4\times4\times2 = 81 - 32\\\Delta = 49\\t=\dfrac{9\pm\sqrt{49}}{2\times 4}\\t_1=\frac{1}{4}\\t_2=2\\2^x=\dfrac{1}{4}\\2^x=\dfrac{1}{2^2}=2^{-2}\\x=-2\\\\ou\\\\2^x=2=2^1\\x=1$

b)

$3^{x-1}-3^x+3^{x+1}+3^{x+2}=306\\3^x\times3^{-1}-3^x+3^x\times3+3^x\times3^2=306\\(3^{-1}-1+3+3^2)\times3^x=306\\(\dfrac{1}{3}+11)\times3^x=306\\\dfrac{34}{3}\times 3^x=306\\\\\therefore 3^x=27=3^3\\x=3$