Question

4^(х^2+5х-4)+3*2^(х^2+5х-4)-4=0

Answer 1

2^(x²+5x-4)=a
a²+3a-4=0
a1+a2=-3 U a1*a2=-4
a1=-4⇒2^(x²+5x-4)=-4 нет решения
a2=1⇒2^(x²+5x-4)=1
x²+5x-4=0
D=25+16=41
x1=(-5-√41)/2 U x2=(-5+√41)/2

Answer 2

To solve the equation 4^(x^2+5x-4) + 3*2^(x^2+5x-4) - 4 = 0, we can substitute a variable to simplify the equation. Let's substitute y = x^2 + 5x - 4.

Now the equation becomes:

4^y + 3*2^y - 4 = 0

We can solve this equation by applying the substitution.

Let's rewrite the equation:

4^y + 3*2^y = 4

Now, we can write 4 as a combination of powers of 2:

2^2 = 4

Substituting this in the equation:

(2^2)^y + 3*2^y = 2^2

Using the properties of exponents, we can simplify the equation further:

2^(2y) + 3*2^y = 2^2

Now we have a common base (2), so we can set the exponents equal to each other:

2y + 3*2^y = 2^2

Rearranging the terms:

2y = 2^2 - 3*2^y

2y = 4 - 3*2^y

2y = 4 - 3*2^y

Now we can solve this equation for y. However, it's not possible to find an exact algebraic solution for this equation. We can use numerical methods or approximations to find an estimate for y.

One popular numerical method is the Newton-Raphson method. By applying this method, we can iteratively refine an initial guess for y until we find a solution that satisfies the equation.

Please provide an initial guess for y, and I can demonstrate the Newton-Raphson method to approximate the solution.