Question

2*log 4(4+x)=4-log2(x-2)

Answer 1

4-4=0 .о4-4 4-4 4-4 4-4 -4 -4м-4 маловато надо на 6 меньше

Answer 2

We can solve the equation using algebraic manipulations.

First, we can simplify the left-hand side of the equation using the logarithmic identity log_a(b^c) = c*log_a(b):

2*log_4(4+x) = log_4((4+x)^2)

Next, we can simplify the right-hand side of the equation by using the logarithmic identity log_a(b/c) = log_a(b) - log_a(c):

4 - log_2(x-2) = log_2(16) - log_2(x-2) = log_2(16/(x-2))

Substituting these simplifications back into the original equation, we get:

log_4((4+x)^2) = log_2(16/(x-2))

To solve for x, we can use the property that if two logarithms with the same base are equal, then their arguments must be equal. Thus:

(4+x)^2 = 4/(x-2)

Expanding the left-hand side and simplifying, we get:

x^2 + 8x + 16 = 4/(x-2)

Multiplying both sides by (x-2), we get:

x^3 - 6x^2 - 8x + 32 = 0

We can use synthetic division to factor the polynomial:

Copy code
2  |  1  -6  -8  32
   |     2  -8  0
   |  1  -4  -16
   

Thus, the polynomial factors as:

(x - 2)(x^2 - 4x - 16) = 0

Using the quadratic formula to solve for the roots of the quadratic factor, we get:

x = 2, 2 + 2sqrt(5), 2 - 2sqrt(5)

However, we need to check which solutions satisfy the original equation, as some solutions may result in undefined or extraneous values.

For x = 2, we have:

2log_4(4+2) = 2log_4(6) = 2*log(6)/log(4) ≈ 2.79

4 - log_2(2-2) = 4

Thus, x = 2 is a valid solution.

For x = 2 + 2*sqrt(5), we have:

2log_4(4+2+2sqrt(5)) ≈ 3.34

4 - log_2(2+2*sqrt(5)-2) ≈ -0.86

Thus, x = 2 + 2*sqrt(5) is not a valid solution.

For x = 2 - 2*sqrt(5), we have:

2log_4(4+2-2sqrt(5)) ≈ -0.73

4 - log_2(2-2*sqrt(5)-2) ≈ -4.61

Thus, x = 2 - 2*sqrt(5) is not a valid solution.

Therefore, the only valid solution is x = 2.