Question

2log2 (x-2)+log 0,5 (x-3)>2​

Answer 1

Решение см на листочке.

Answer 2

To solve the inequality 2log2(x-2) + log0.5(x-3) > 2, we need to use logarithmic properties to simplify the expression and then find the solution for x.

Logarithmic properties:

1. log_a(b) + log_a(c) = log_a(b * c)
2. log_a(b^n) = n * log_a(b)

Using these properties, we can simplify the given inequality:

2log2(x-2) + log0.5(x-3) > 2

Apply property 1 to combine the logarithms:

log2((x-2)^2) + log0.5(x-3) > 2

Apply property 2 to the first logarithm:

log2((x-2)^2) = 2 * log2(x-2)

Now the inequality becomes:

2 * log2(x-2) + log0.5(x-3) > 2

Now, convert the logarithms to the same base:

log0.5(x-3) = log2((x-3)^(-1))

The inequality becomes:

2 * log2(x-2) + log2((x-3)^(-1)) > 2

Combine the logarithms using property 1:

log2((x-2)^2) + log2((x-3)^(-1)) > 2

Now apply another property to combine the logarithms:

log2((x-2)^2 * (x-3)^(-1)) > 2

Now, remove the logarithm using the definition of logarithms:

(x-2)^2 * (x-3)^(-1) > 2^2

Simplify the right-hand side:

(x-2)^2 * (x-3)^(-1) > 4

Now, multiply both sides by (x-3) to get rid of the denominator:

(x-2)^2 > 4(x-3)

Expand the left side:

x^2 - 4x + 4 > 4x - 12

Move all terms to one side of the inequality:

x^2 - 8x + 16 > 0

Now we need to find the critical points of the inequality, where the expression equals zero:

x^2 - 8x + 16 = 0

Using the quadratic formula:

x = (8 ± √(8^2 - 4 * 1 * 16)) / 2

x = (8 ± √(64 - 64)) / 2

x = (8 ± √0) / 2

x = 8 / 2

x = 4

Now, we have the critical point x = 4. We can now analyze the inequality in different intervals:

1. x < 4
2. x = 4
3. x > 4

Test the intervals using a test point in each one:

1. Choose x = 3 (a number less than 4):

(x-2)^2 > 4(x-3) (3-2)^2 > 4(3-3) 1 > 0 (True)

2. Choose x = 4 (the critical point):

(x-2)^2 > 4(x-3) (4-2)^2 > 4(4-3) 4 > 4 (False)

3. Choose x = 5 (a number greater than 4):

(x-2)^2 > 4(x-3) (5-2)^2 > 4(5-3) 9 > 8 (True)

The inequality is true for x < 4 and x > 4. However, since x = 4 is not included in the solution (due to the strict inequality sign), the final solution is:

x < 4 or x > 4