Question

Log2(3x+1)>log2(2-7x)

Answer 1

log2(3x-1)(2-7x)>1

21x^2-13x+2=0

D=1;x1=1/3;x2=2/7

Answer 2

To solve the inequality log2(3x + 1) > log2(2 - 7x), we need to apply logarithm properties and algebraic techniques. Remember that logarithms are only defined for positive values, so we need to ensure that the arguments inside the logarithms are positive. Here's the step-by-step solution:

Step 1: Ensure that the arguments inside the logarithms are positive. Since both arguments are expressions that can take any real value, we need to consider two cases where each argument is positive:

Case 1: 3x + 1 > 0 Solve for x: 3x + 1 > 0 3x > -1 x > -1/3

Case 2: 2 - 7x > 0 Solve for x: 2 - 7x > 0 -7x > -2 x < 2/7

So the valid range for x is: -1/3 < x < 2/7

Step 2: Compare the expressions inside the logarithms.

log2(3x + 1) > log2(2 - 7x)

Step 3: Apply logarithm properties.

We can use the logarithm property: If loga(b) > loga(c), then b > c.

Using this property, we get:

3x + 1 > 2 - 7x

Step 4: Solve the inequality.

Add 7x to both sides:

3x + 7x + 1 > 2

Combine like terms:

10x + 1 > 2

Subtract 1 from both sides:

10x > 1

Divide by 10:

x > 1/10

Step 5: Check the validity of the solution.

Now that we have solved the inequality, let's check if the solution x > 1/10 is within the valid range we found earlier: -1/3 < x < 2/7.

1/10 is within the valid range, so the final solution to the inequality is:

x > 1/10, or in interval notation, (1/10, 2/7).