Question

25^x-6*5^x+5=0 equação exponencial​

Answer 1

Resposta:

S={0,1}

Explicação passo-a-passo:

$\displaystyle\\25^{x}+6.5^{x}+5=0\\5^{2x}+6.5^{x}+5=0\\5^{x}=y\\y^{2}+6y+5=0\\\\Aplicando~a~f\'{o}rmula~de~Bhaskara~para~y^{2}-6y+5=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-6~e~c=5\\\\\Delta=(b)^{2}-4(a)(c)=(-6)^{2}-4(1)(5)=36-(20)=16\\\\y^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-6)-\sqrt{16}}{2(1)}=\frac{6-4}{2}=\frac{2}{2}=1\\\\y^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-6)+\sqrt{16}}{2(1)}=\frac{6+4}{2}=\frac{10}{2}=5\\\\Para~y=1\\5^{x}=y\\5^{x}=1\\5^{x}=5^{0}\\x=0$

$Para~y=5\\5^{x}=y\\5^{x}=5\\5^{x}=5^{1}\\x=1$

Answer 2

Resposta: S = (0, 1)

25^x - 6*5^x + 5 = 0 equação exponencial​

Explicação passo-a-passo:

y = 5^x

y^2 - 6y + 5 = 0

os coeficientes

a = 1, b = -6, c = 5

delta

d = 36 - 20 = 16

y1 = ( 6 + 4 ) / 2 = 10 / 2 = 5

5^x1 = 5^1, x1 = 1

y2 = (6 - 4 ) / 2 = 1

5^x2 = 1 = 5^0 , x2 = 0

S = (0, 1)