Question

4^(x+1)-9×2^(x+1)+8=0

Answer 1

4^(x+1)-9*2^(x+1)+8=0

2^2(x+1)-9*2^(x+1)+8=0

2^(x+1)=t,t>0

t²-9t+8=0

D=81-32=49=7²

t1=9+7/2=8

t2=9-7/2=1

Теперь подставляем место t:

2^(x+1)=t1

2^(x+1)=8

2^(x+1)=2³

x+1=3

x=4-является

2^(x+1)=t2

2^(x+1)=1

2^(x+1)=2^0

x+1=0

x=-1-является

Ответ:4,-1;

Одз.
х+1≥0

----*--->
-1

Answer 2

To solve the equation 4^(x+1) - 9*2^(x+1) + 8 = 0, we can use some algebraic manipulation and techniques. Let's go step by step:

1. Notice that 4^(x+1) can be written as (2^2)^(x+1) = 2^(2x+2).
2. Similarly, 92^(x+1) can be written as 9(2^1)^(x+1) = 9*2^(x+2).

Now, our equation becomes:

2^(2x+2) - 9*2^(x+2) + 8 = 0.

3. Let's make a substitution to simplify the equation. Let's introduce a new variable: y = 2^(x+2).

Now, our equation becomes:

y^2 - 9y + 8 = 0.

4. Factor the quadratic equation:

(y - 8)(y - 1) = 0.

5. Now, solve for y:

y - 8 = 0 -> y = 8, y - 1 = 0 -> y = 1.

6. Remember that y = 2^(x+2), so we can now solve for x in each case:

For y = 8: 2^(x+2) = 8, 2^(x+2) = 2^3.

Since the bases are the same, we equate the exponents:

x + 2 = 3, x = 3 - 2, x = 1.

For y = 1: 2^(x+2) = 1, 2^(x+2) = 2^0.

Again, equate the exponents:

x + 2 = 0, x = 0 - 2, x = -2.

So the solutions for the equation are x = 1 and x = -2.