Question

Se x e y são reais tais que $ln[(y^2 +10).e^x] - ln(y^2 +1)^4= x-3$  então:

a)$1+ \sqrt{e-1}$

b)$10 - \sqrt{e-1}$

c)$+- \sqrt{e-1}$

d)$+- \sqrt{e+1}$

e)$\sqrt{e- \frac{1}{2} }$

Answer 1

$ Ola Matheus

ln((y² + 1)*e^x) - ln(y² + 1)^4 = x - 3

x + ln(y² + 1) - ln(y² + 1)^4 = x - 3

ln(y² + 1)^4 - ln(y² + 1) = 3

ln(y² + 1)^3 = 3

3ln(y² + 1) = 3

ln(y² + 1) = 1

y² + 1 = e

y² = e - 1

y = ± √(e - 1) 

pronto

Answer 2

$ln[y^2+1)e^x-ln(y^2+1)^4=x-3 =\ \textgreater \ ln(y^2+1) + lne^x-ln(y^2+1)^4 \\ ln \frac{y^2+1}{(y^2+1)^4}+x.lne=x-3 =\ \textgreater \ ln \frac{1}{(y^2+1)^3}+x . 1 =x-3 \\ ln \frac{1}{(y^2+1)^3} =-3=\ \textgreater \ \frac{1}{(y^2+1)^3} =e^-^3 \\ =\ \textgreater \ \frac{1}{(y^2+1)^3} = \frac{1}{e^3} =\ \textgreater \ (y^2+1)^3=e^3=\ \textgreater \ y^2+1=e=\ \textgreater \ y^2=e-1 \\y= \sqrt{e-1} =\ \textgreater \ y= \sqrt{2,718-1} =\ \textgreater \ y=(+-)1,31 \\ \\ \\ b$