Math Problem Statement

Solve the inequality log_2(2x - 1) < log_(1/2)(1 - x).

Solution

The inequality in the image is:

log2(2x1)<log12(1x)\log_2(2x - 1) < \log_{\frac{1}{2}}(1 - x)

To solve this inequality, let's go step-by-step.

Step 1: Rewrite the logarithms

We will convert both logarithmic expressions to base 2. We know that:

log12(a)=log2(a)log2(12)\log_{\frac{1}{2}}(a) = \frac{\log_2(a)}{\log_2\left(\frac{1}{2}\right)}

Since log2(12)=1\log_2\left(\frac{1}{2}\right) = -1, we can rewrite the second term as:

log12(1x)=log2(1x)\log_{\frac{1}{2}}(1 - x) = -\log_2(1 - x)

Step 2: Set up the new inequality

Now, the inequality becomes:

log2(2x1)<log2(1x)\log_2(2x - 1) < -\log_2(1 - x)

Step 3: Eliminate the negative sign

We can multiply both sides of the inequality by 1-1 (remember that this reverses the inequality sign):

log2(2x1)>log2(1x)-\log_2(2x - 1) > \log_2(1 - x)

Or equivalently:

log2(12x1)>log2(1x)\log_2\left(\frac{1}{2x - 1}\right) > \log_2(1 - x)

Step 4: Remove the logarithms

Since log2(a)>log2(b)\log_2(a) > \log_2(b) implies that a>ba > b, we can eliminate the logarithms to get:

12x1>1x\frac{1}{2x - 1} > 1 - x

Step 5: Solve the inequality

Now, we need to solve the inequality 12x1>1x\frac{1}{2x - 1} > 1 - x. Let's first multiply both sides by 2x12x - 1 (note that 2x12x - 1 must be positive for this to be valid, so we need to check the domain later):

1>(1x)(2x1)1 > (1 - x)(2x - 1)

Now expand the right-hand side:

1>2x12x2+x1 > 2x - 1 - 2x^2 + x

Simplifying:

1>2x2+3x11 > -2x^2 + 3x - 1

Rearrange the inequality:

2x23x+2>02x^2 - 3x + 2 > 0

This is a quadratic inequality. We will solve it by finding the roots of the corresponding equation 2x23x+2=02x^2 - 3x + 2 = 0 and determining the intervals.

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Math Problem Analysis

Mathematical Concepts

Logarithmic Inequalities
Logarithm Base Conversion
Quadratic Inequalities

Formulas

log_a(b) = log_c(b) / log_c(a)
log_(1/2)(x) = -log_2(x)
Quadratic equation: ax^2 + bx + c = 0

Theorems

Logarithmic Properties
Quadratic Formula

Suitable Grade Level

High School (Grades 10-12)