Question

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

Answer 1

x>3+2log2(5) (двва после log - маленькая)

Answer 2

The given inequality can be rewritten using logarithmic rules as:

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

Now we can use the properties of logarithms to simplify the equation. Recall that:

• loga(b) + loga(c) = loga(bc)
• loga(b) - loga(c) = loga(b/c)
• loga(b^c) = cloga(b)

Using these properties, we can simplify the left-hand side of the inequality as follows:

2log2(x-2) + log0.5(x-3) = log2[(x-2)^2] + log0.5(x-3) [using the first property] = log2[(x-2)^2 * 0.5(x-3)] [using the second property] = log2[(x-2)^2 * (x-3)/2]

Now the inequality becomes:

log2[(x-2)^2 * (x-3)/2] > 2

Using the definition of logarithms, we can rewrite this as:

2^(log2[(x-2)^2 * (x-3)/2]) > 2^2

Simplifying the left-hand side using the third logarithmic property, we get:

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

Multiplying both sides by 2 to eliminate the fraction, we get:

(x-2)^2 * (x-3) > 8

Expanding the left-hand side, we get:

x^3 - 7x^2 + 16x - 12 > 0

Now we need to find the values of x that satisfy this inequality. We can use a sign chart to do this:

markdownCopy code
    x      x^3 - 7x^2 + 16x - 12
---------------------------------
  -inf             -    
  2-1.28           +    
  1.28-3           -    
  3-2              +    
   2              -    
  2+inf            +    

The inequality is satisfied when x < 1.28 or x > 3. Therefore, the solution to the original inequality is:

x < 1.28 or x > 3

Answer 3

The inequality can be rewritten using logarithm rules:

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

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

log2[(x-2)^2 * 0.5(x-3)] > 2

Now we can rewrite the logarithmic inequality as an exponential inequality:

2[(x-2)^2 * 0.5(x-3)] > 2^2

(x-2)^2 * 0.5(x-3) > 2

Multiplying both sides by 2 to get rid of the fraction:

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

Expanding the left side:

x^3 - 7x^2 + 16x - 12 > 0

Now we can solve the inequality by finding the roots of the polynomial and testing the sign of each interval between the roots:

(x-2)(x-2)(x-4) > 0

The roots are x = 2, x = 2, and x = 4. We have a repeated root at x=2.

Testing the sign of each interval:

Interval (-∞, 2):

Choose x=-1, then x^3 - 7x^2 + 16x - 12 = -30, which is negative.

Interval (2, 4):

Choose x=3, then x^3 - 7x^2 + 16x - 12 = 10, which is positive.

Interval (4, ∞):

Choose x=5, then x^3 - 7x^2 + 16x - 12 = 18, which is positive.

Therefore, the solution to the inequality is:

x < 2 or x > 4.

Note that x=2 is not included in the solution since it would make the logarithm undefined.