Question

8 * 2^2x - 2 * 2^2x=0

Answer 1

$8\cdot 2^{2x}-2\cdot 2^{2x}=0\\\\6\cdot 2^{2x}=0\\\\2^{2x}=0\; \; \Rightarrow \; \; \boxed {x\in \varnothing }\; \; ,\; \; t.k.\; \; 2^{2x}>0\; (2^{2x}\ne 0\; )\; \; pri\; \; x\in R.$

Answer 2

To solve the equation 8 * 2^(2x) - 2 * 2^(2x) = 0, you can factor out a common term, which is 2^(2x):

2^(2x) * (8 - 2) = 0

Now, simplify:

2^(2x) * 6 = 0

To solve for x, set each factor equal to zero:

1. 2^(2x) = 0
2. 6 = 0

The second equation, 6 = 0, is never true, so it has no solutions.

Now, let's focus on the first equation, 2^(2x) = 0. This equation cannot be satisfied because 2 raised to any power will always be greater than zero. Therefore, there are no solutions for x in this equation.

In conclusion, the original equation 8 * 2^(2x) - 2 * 2^(2x) = 0 has no solutions for x.

Answer 3

To solve the equation 8 * 2^(2x) - 2 * 2^(2x) = 0, you can factor out a common term:

2^(2x) * (8 - 2) = 0

Now, simplify further:

2^(2x) * 6 = 0

Now, you have:

2^(2x) = 0/6

Since anything divided by 6 is still 0, you have:

2^(2x) = 0

However, 2 to any power (positive or negative) is always greater than 0, so there are no real solutions to this equation. In other words, there is no value of x that will make the equation true.