Question

2log2(x)=1+log2(x+4)

Answer 1

2㏒₂x=㏒₂2+㏒₂(x+4)
ОДЗ: х>0
x²=2(x+4)
x²-2x-8=0
D=36
x₁=4; x₂=-2
Ответ: 4.

Answer 2

To solve this equation for x, we can begin by simplifying it using logarithmic rules. Specifically, we can use the rule that says log(a^n) = nlog(a) and the rule that says log(ab) = log(a) + log(b). Applying these rules, we get:

2log2(x) = 1 + log2(x+4) log2(x^2) = log2(2) + log2(x+4) log2(x^2) = log2(2(x+4)) log2(x^2) = log2(2x+8)

Now, we can use the fact that if loga(b) = loga(c), then b = c. Applying this to our equation, we get:

x^2 = 2x + 8

This is a quadratic equation that we can solve by moving all the terms to one side of the equation and using the quadratic formula:

x^2 - 2x - 8 = 0 x = [2 ± sqrt(4 + 4*8)] / 2 x = [2 ± 6] / 2

So, the solutions to the equation are x = 4 or x = -2. However, we need to check if these solutions are valid by plugging them back into the original equation and verifying that they satisfy the equation.

If we plug in x = 4, we get:

2log2(4) = 1 + log2(4+4) 2(2) = 1 + 3 4 = 4

This solution checks out. If we plug in x = -2, we get:

2log2(-2) = 1 + log2(-2+4) This is undefined since log2 of a negative number is undefined.

Therefore, the only solution to the equation is x = 4.