Question

25^x + 3 * 10^x - 4 * 4^x=0

Answer 1

Ответ:

Объяснение:

На фото

Answer 2

To solve the equation 25^x + 3 * 10^x - 4 * 4^x = 0, we can rewrite it in terms of a common base. Let's express all the terms using the base 5, since 25 = 5^2 and 10 = 5 * 2:

(5^2)^x + 3 * (5 * 2)^x - 4 * (2^2)^x = 0

Simplifying further:

5^(2x) + 3 * 5^x * 2^x - 4 * 2^(2x) = 0

Now, let's substitute a variable to make the equation easier to work with. Let y = 5^x:

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

This quadratic equation can be factored as:

(y - 4 * 4^x)(y + 4 * 4^x + 4) = 0

Now we have two cases to consider:

Case 1: y - 4 * 4^x = 0

Substituting y = 5^x back in:

5^x - 4 * 4^x = 0

Dividing both sides by 4^x:

(5/4)^x = 1

Since any number raised to the power of 0 is equal to 1, we have:

(5/4)^x = 1

Taking the logarithm of both sides with base 5/4:

x * log(5/4) = log(1)

Since log(1) = 0, the equation simplifies to:

x * log(5/4) = 0

This means that x can be any real number.

Case 2: y + 4 * 4^x + 4 = 0

Substituting y = 5^x back in:

5^x + 4 * 4^x + 4 = 0

We can rewrite 4 as 2^2:

5^x + 4 * 2^(2x) + 4 = 0

Now we can substitute z = 2^x:

5^x + 4 * z^2 + 4 = 0

This quadratic equation can be solved using the quadratic formula:

z = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 4, b = 0, and c = 4. Substituting these values into the quadratic formula:

z = (-0 ± √(0^2 - 4 * 4 * 4)) / (2 * 4) z = (± √(-64)) / 8

Since the term inside the square root is negative, the equation has no real solutions in this case.

To summarize, the equation 25^x + 3 * 10^x - 4 * 4^x = 0 has solutions where x can be any real number, as well as cases where there are no real solutions.