Question

Log^22(вторая двойка твнизу)-4log2x=12

Answer 1

$log(2)^{2} - 4 log(2x) = 12 \\ x > 0 \\ - 4 log(2x) = 12 - log(2)^{2} \\ log(2x) = \frac{12 - log(2)^{2} }{ - 4} \\ log(2x) = - 3 + \frac{ { log(2) }^{2} }{4} \\ 2x = 10^{ - 3 + \frac{ { log(2) }^{2} }{4} } \\ x = \frac{10^{ - 3 + \frac{ { log(2) }^{2} }{4} } }{2}$

Answer 2

To solve the equation, we'll first need to clarify the expression "Log^22(вторая двойка твнизу)" as it is not clear what it represents. However, I'll assume that "вторая двойка твнизу" means the number 2, and "Log^22" means the logarithm base 2.

With this assumption, the equation becomes:

log2(2) - 4 log2(x) = 12

We can simplify this as follows:

1 - 4 log2(x) = 12

Now, let's isolate the logarithmic term:

-4 log2(x) = 12 - 1

-4 log2(x) = 11

Next, divide both sides by -4 to solve for log2(x):

log2(x) = -11/4

Now, to find the value of x, we'll use the definition of logarithms:

x = 2^(log2(x))

x = 2^(-11/4)

x ≈ 0.0335

So, the value of x that satisfies the equation is approximately 0.0335.