Recent content by Avalance789

  1. A

    MHB F(x)=ab^{x} where b must be a positive real number

    Ok, got it. So from this point we get into complex numbers. Thank you! Отправлено с моего SM-A750FN через Tapatalk
  2. A

    MHB F(x)=ab^{x} where b must be a positive real number

    It means sqrt{-1}? Отправлено с моего SM-A750FN через Tapatalk
  3. A

    MHB F(x)=ab^{x} where b must be a positive real number

    Quote: "In mathematics, an exponential function is a function of the form f(x)=ab^{x} where b is a POSITIVE REAL number" Wait. Give me a reason, why exponent base must be positive and real? What happens if b<0?
  4. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    Ок, but if a=1, then x=1. Cannot be that -5/3<a<5/3
  5. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    Just tell me if a<2/3 is correct answer. I will be happiest person on Earth, guys
  6. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    And after that simplification?
  7. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    4x^2-a^2 - 2x - a = 0 So we need to find a values when D=0? And if a=1, then it's quadrant equation?
  8. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    Sorry, I have confused you. Should be written like sqrt (5-6x)*ln(4x^2-a^2)=sqrt(5-6x)*ln(2x+a)So sorry
  9. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    Teacher gives me result (-5/3; -1/2] [2/3; 5/3) Is it correct?
  10. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    Forces me to solve it ultimatively
  11. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    So it doesn't have solution? My teacher states on that it has
  12. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    I am sorry, but I am unable to determine it. But if I won't solve it, I am dead man(
  13. A

    MHB √(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

    Sqrt (5-6*x)*ln(4*sqr(x)-sqr(a))=sqrt(5-6*x)*ln(2*x+a) Find all possible a when an equation has only one possible solution.