Question

4^x-15*2^x+16=0 найти x1, x2

Answer 1

Ответ:

Пошаговое объяснение:

$4^x-15*2^x+16=0\\2^{2x}-15*2^x+16=0\\t=2^x\\t^2-15t+16=0\\\\D=b^2-4ac=225-4*16=225-64=161\\t1=\frac{15+\sqrt{161} }{2} \\\\t2=\frac{15-\sqrt{161} }{2}\\\\2^x=\frac{15+\sqrt{161} }{2}\\2^x=\frac{15-\sqrt{161} }{2}\\\\x1=log{2}(15+\sqrt{161} )-1\\x1=log{2}(15-\sqrt{161} )-1$

Answer 2

To solve the equation 4^x - 15*2^x + 16 = 0 for x, we can use substitution to simplify the equation and then solve for x. Let's make a substitution:

Let y = 2^x.

Now the equation becomes:

4^x - 152^x + 16 = 0 (2^2)^x - 152^x + 16 = 0 y^2 - 15y + 16 = 0

Now, we have a quadratic equation in terms of y. We can factor it or use the quadratic formula to find the values of y.

The quadratic equation y^2 - 15y + 16 = 0 can be factored as:

(y - 1)(y - 16) = 0

Now, set each factor equal to zero and solve for y:

1. y - 1 = 0 y = 1

2. y - 16 = 0 y = 16

Now, remember that y = 2^x, so we can solve for x in each case:

1. y = 1 2^x = 1 x = 0

2. y = 16 2^x = 16 x = log2(16) x = 4

So, the solutions to the equation are x1 = 0 and x2 = 4.